面试算法大全-树与二叉树总结

6、树结构

6.1 树数据结构体系

6.1.1 二叉树和基本实现

二叉树数据结构定义：

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    public TreeNode() {
    }

    public TreeNode(int val) {
        this.val = val;
    }

    public TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }

    public int getVal() {
        return val;
    }

    public void setVal(int val) {
        this.val = val;
    }

    public TreeNode getLeft() {
        return left;
    }

    public void setLeft(TreeNode left) {
        this.left = left;
    }

    public TreeNode getRight() {
        return right;
    }

    public void setRight(TreeNode right) {
        this.right = right;
    }
}

二叉树的前序，中序，后序，层序遍历(递归)

/**
     * 先序遍历
     *
     * @param node
     */
public void preOrder(Node node) {
    if (node == null) {
        return;
    }
    System.out.print(node.val);
    preOrder(node.left);
    preOrder(node.right);
}
/**
     * 中序遍历
     *
     * @param node
     */
public void midOrder(Node node) {
    if (node == null) {
        return;
    }
    midOrder(node.left);
    System.out.print(node.val);
    midOrder(node.right);
}
/**
     * 后序遍历
     *
     * @param node
     */
public void postOrder(Node node) {
    if (node == null) {
        return;
    }
    postOrder(node.left);
    postOrder(node.right);
    System.out.print(node.val);
}

二叉树的前序，中序，后序，层序遍历(非递归)

/**
     * 前序遍历
     * 非递归形式
     *
     * @param root
     */
private void preOrderStack(Node root) {
    if (root == null) {
        return;
    }
    Stack<Node> stack = new Stack<>();
    stack.push(root);
    while (!stack.isEmpty()) {
        Node node = stack.pop();
        System.out.print(node.val);
        if (node.right != null) {
            stack.push(node.right);
        }
        if (node.left != null) {
            stack.push(node.left);
        }
    }
}
/**
     * 中序遍历，非递归形式
     *
     * @param root
     */
private void midOrderStack(Node root) {
    if (root == null) {
        return;
    }
    Stack<Node> stack = new Stack<>();
    //找一个辅助节点，借助辅助节点寻找最左节点。
    Node cur = root;
    while (!stack.isEmpty() || cur != null) {
        while (cur != null) {
            stack.push(cur);
            cur = cur.left;
        }
        Node node = stack.pop();
        System.out.print(node.val);
        if (node.right != null) {
            cur = node.right;
        }
    }
}
/**
     * 后序遍历 ，非递归形式
     *
     * @param root
     */
private void postOrderStack(Node root) {
    if (root == null) {
        return;
    }
    Stack<Node> stack = new Stack<>();
    Stack<Node> stack1 = new Stack<>();
    stack.push(root);
    while (!stack.isEmpty()) {
        Node node = stack.pop();
        stack1.push(node);
        if (node.left != null) {
            stack.push(node.left);
        }
        if (node.right != null) {
            stack.push(node.right);
        }
    }
    while (!stack1.isEmpty()) {
        System.out.print(stack1.pop().val);
    }
}

二叉树的层序遍历

/**
     * 层序遍历
     * @param root
     */
private void bfs(Node root) {
    if (root == null) {
        return;
    }
    Queue<Node> queue = new LinkedList<>();
    queue.offer(root);
    while (!queue.isEmpty()){
        Node node=queue.poll();
        System.out.print(node.val);
        if (node.left!=null){
            queue.offer(node.left);
        }
        if (node.right!=null){
            queue.offer(node.right);
        }
    }
}
/**
     * 层序遍历,保存每一层遍历的值
     * @param root
     */
private List<List<String>> bfs2(Node root){
    List<List<String>> res=new ArrayList<>();
    if (root==null){
        return res;
    }
    Queue<Node> queue=new LinkedList<>();
    queue.offer(root);
    List<String> tmp;
    while (!queue.isEmpty()){
        int size = queue.size();
        tmp=new ArrayList<>();
        while (size-- >0){
            Node node=queue.poll();
            tmp.add(node.val);
            if (node.left!=null){
                queue.add(node.left);
            }
            if (node.right!=null){
                queue.add(node.right);
            }
        }
        res.add(tmp);
    }
    return  res;
}

二叉搜索树具有有以下性质的二叉树：
  （1）若左子树不为空，则左子树上所有节点的值均小于或等于它的根节点的值。
  （2）若右子树不为空，则右子树上所有节点的值均大于或等于它的根节点的值。
  （3）左、右子树也分别为二叉搜索树。

/**
 * Java实现二叉查找树
 * @author Kiritor
 * @param <T>*/
public class BinarySearchTree<T extends Comparable<? super T>> {
    /**结点数据结构*/
    static class BinaryNode<T>
    {
        T data;
        BinaryNode<T> left;
        BinaryNode<T> right;
        public BinaryNode(T data) {
            this(data,null,null);
        }
        public BinaryNode( T data, BinaryNode<T> left, BinaryNode<T> right) {
            this.data =data;
            this.left = left;
            this.right =right;
        }
        public BinaryNode()
        {
            data =null;
            this.left = left;
            this.right =right;
        }
    }
    private BinaryNode<T> rootTree;
    /**构造一颗空的二叉查找树*/
    public BinarySearchTree()
    {
        rootTree = null;
    }
    /**清空二叉查找树*/
    public void clear()
    {
        rootTree = null;
    }
    /**判断是否为空*/
    public boolean isEmpty()
    {
        return rootTree == null;
    }
    /**查找指定的元素,默认从
     * 根结点出开始查询*/
    public boolean contains(T t)
    {
        return contains(t, rootTree);

    }
    /**找到二叉查找树中的最小值*/
    public T findMin()
    {
        if(isEmpty())
        {
            System.out.println("二叉树为空");
            return null;
        }else {
            return findMin(rootTree).data;
        }

    }
    /**找到二叉查找树中的最大值*/
    public T findMax()
    {
        if(isEmpty())
        {
            System.out.println("二叉树为空");
            return null;
        }else {
            return findMax(rootTree).data;
        }
    }
    /**插入元素*/
    public void insert(T t)
    {
        rootTree = insert(t, rootTree);
    }
    /**删除元素*/
    public void remove(T t)
    {
        rootTree = remove(t,rootTree);
    }
    /**打印二叉查找树*/
    public void printTree()
    {

    }
    /**从某个结点出开始查找元素*/
    public boolean contains(T t, BinaryNode<T> node)
    {
        if(node==null) {
            return false;
        }
        int result = t.compareTo(node.data);
        if(result>0) {
            return contains(t,node.right);
        } else if(result<0) {
            return contains(t, node.left);
        } else {
            return true;
        }
    }
    /**查询出最小元素所在的结点*/
    public BinaryNode<T> findMin(BinaryNode<T> node)
    {
        if(node==null) {
            return null;
        } else if(node.left==null) {
            return node;
        }
        return findMin(node.left);//递归查找
    }
    /**查询出最大元素所在的结点*/
    public BinaryNode<T> findMax(BinaryNode<T> node)
    {
        if(node!=null)
        {
            while(node.right!=null) {
                node=node.right;
            }
        }
        return node;
    }
    /**在某个位置开始判断插入元素*/
    public BinaryNode<T> insert(T t,BinaryNode<T> node)
    {
        if(node==null)
        {
            //新构造一个二叉查找树
            return new BinaryNode<T>(t, null, null);
        }
        int result = t.compareTo(node.data);
        if(result<0) {
            node.left= insert(t,node.left);
        } else if(result>0) {
            node.right= insert(t,node.right);
        } else {
            ;//doNothing
        }
        return node;
    }
    /**在某个位置开始判断删除某个结点*/
    public BinaryNode<T> remove(T t,BinaryNode<T> node)
    {
        if(node == null) {
            return node;//没有找到,doNothing
        }
        int result = t.compareTo(node.data);
        if(result>0) {
            node.right = remove(t,node.right);
        } else if(result<0) {
            node.left = remove(t,node.left);
        } else if(node.left!=null&&node.right!=null)
        {
            node.data = findMin(node.right).data;
            node.right = remove(node.data,node.right);
        }
        else {
            node = (node.left!=null)?node.left:node.right;
        }
        return node;

    }
    public BinaryNode<Integer> init()
    {
        BinaryNode<Integer> node3 = new BinaryNode<Integer>(3);
        BinaryNode<Integer> node1 = new BinaryNode<Integer>(1);
        BinaryNode<Integer> node4 = new BinaryNode<Integer>(4,node3,null);
        BinaryNode<Integer> node2 = new BinaryNode<Integer>(2,node1,node4);
        BinaryNode<Integer> node8 = new BinaryNode<Integer>(8);
        BinaryNode<Integer> root = new BinaryNode<Integer>(6,node2,node8);
        return root;
    }
    public void preOrder(BinaryNode node) {
        if (node != null) {
            System.out.print(node.data);
            preOrder(node.left);
            preOrder(node.right);
        }
    }
    /*简单测试*/
    public static void main(String[] args) {
        BinarySearchTree  searchTree = new BinarySearchTree<>();
        BinaryNode<Integer> node= searchTree.init();
        searchTree.rootTree=node;
        searchTree.preOrder(searchTree.rootTree);
        searchTree.remove(4);
        searchTree.preOrder(searchTree.rootTree);
    }
}

平衡二叉树

​ （1）它的结点左子树和右子树的深度之差不超过1

​ （2）该结点的左子树和右子树都是一棵平衡二叉树

/**
 *二叉平衡树简单实现
 *@author kiritor
 */
public class AvlTree< T extends Comparable< ? super T>>
{
    private static class AvlNode< T>{//avl树节点

        AvlNode( T theElement )
        {
            this( theElement, null, null );
        }
        AvlNode( T theElement, AvlNode< T> lt, AvlNode< T> rt )
        {
            element  = theElement;
            left     = lt;
            right    = rt;
            height   = 0;
        }
        T           element;      // 节点中的数据
        AvlNode< T>  left;         // 左儿子
        AvlNode< T>  right;        // 右儿子
        int         height;       // 节点的高度
    }
    private AvlNode< T> root;//avl树根
    public AvlTree( )
    {
        root = null;
    }
    //在avl树中插入数据，重复数据复略
    public void insert( T x )
    {
        root = insert( x, root );
    }
    //在avl中删除数据,这里并未实现
    public void remove( T x )
    {
        System.out.println( "Sorry, remove unimplemented" );
    }
    //在avl树中找最小的数据
    public T findMin( )
    {
        if( isEmpty( ) ) {
            System.out.println("树空");
        }
        ;
        return findMin( root ).element;
    }
    //在avl树中找最大的数据
    public T findMax( )
    {
        if( isEmpty( ) ) {
            System.out.println("树空");
        }
        return findMax( root ).element;
    }
    //搜索
    public boolean contains( T x )
    {
        return contains( x, root );
    }
    public void makeEmpty( )
    {
        root = null;
    }
    public boolean isEmpty( )
    {
        return root == null;
    }
    //排序输出avl树
    public void printTree( )
    {
        if( isEmpty( ) ) {
            System.out.println( "Empty tree" );
        } else {
            printTree( root );
        }
    }
    private AvlNode< T> insert( T x, AvlNode< T> t )
    {
        if( t == null ) {
            return new AvlNode< T>( x, null, null );
        }
        int compareResult = x.compareTo( t.element );
        if( compareResult < 0 )
        {
            t.left = insert( x, t.left );//将x插入左子树中
            if( height( t.left ) - height( t.right ) == 2 )//打破平衡
            {
                if( x.compareTo( t.left.element ) < 0 )//LL型（左左型）
                {
                    t = rotateWithLeftChild( t );
                } else   //LR型（左右型）
                {
                    t = doubleWithLeftChild( t );
                }
            }
        }
        else if( compareResult > 0 )
        {
            t.right = insert( x, t.right );//将x插入右子树中
            if( height( t.right ) - height( t.left ) == 2 )//打破平衡
            {
                if( x.compareTo( t.right.element ) > 0 )//RR型（右右型）
                {
                    t = rotateWithRightChild( t );
                } else                           //RL型
                {
                    t = doubleWithRightChild( t );
                }
            }
        }
        else {
            ;  // 重复数据，什么也不做
        }
        t.height = Math.max( height( t.left ), height( t.right ) ) + 1;//更新高度
        return t;
    }
    //找最小
    private AvlNode< T> findMin( AvlNode< T> t )
    {
        if( t == null ) {
            return t;
        }
        while( t.left != null ) {
            t = t.left;
        }
        return t;
    }
    //找最大
    private AvlNode< T> findMax( AvlNode< T> t )
    {
        if( t == null ) {
            return t;
        }
        while( t.right != null ) {
            t = t.right;
        }
        return t;
    }
    //搜索（查找）
    private boolean contains( T x, AvlNode t )
    {
        while( t != null )
        {
            int compareResult = x.compareTo( (T) t.element );
            if( compareResult < 0 ) {
                t = t.left;
            } else if( compareResult > 0 ) {
                t = t.right;
            } else {
                return true;    // Match
            }
        }
        return false;   // No match
    }
    //中序遍历avl树
    private void printTree( AvlNode< T> t )
    {
        if( t != null )
        {
            printTree( t.left );
            System.out.println( t.element );
            printTree( t.right );
        }
    }
    //求高度 
    private int height( AvlNode< T> t )
    {
        return t == null ? -1 : t.height;
    }
    //带左子树旋转,适用于LL型
    private AvlNode< T> rotateWithLeftChild( AvlNode< T> k2 )
    {
        AvlNode< T> k1 = k2.left;
        k2.left = k1.right;
        k1.right = k2;
        k2.height = Math.max( height( k2.left ), height( k2.right ) ) + 1;
        k1.height = Math.max( height( k1.left ), k2.height ) + 1;
        return k1;
    }
    //带右子树旋转，适用于RR型
    private AvlNode< T> rotateWithRightChild( AvlNode< T> k1 )
    {
        AvlNode< T> k2 = k1.right;
        k1.right = k2.left;
        k2.left = k1;
        k1.height = Math.max( height( k1.left ), height( k1.right ) ) + 1;
        k2.height = Math.max( height( k2.right ), k1.height ) + 1;
        return k2;
    }
    //双旋转，适用于LR型
    private AvlNode< T> doubleWithLeftChild( AvlNode< T> k3 )
    {
        k3.left = rotateWithRightChild( k3.left );
        return rotateWithLeftChild( k3 );
    }
    //双旋转,适用于RL型
    private AvlNode< T> doubleWithRightChild( AvlNode< T> k1 )
    {
        k1.right = rotateWithLeftChild( k1.right );
        return rotateWithRightChild( k1 );
    }
    // Test program
    public static void main( String [ ] args )
    {
        AvlTree< Integer> t = new AvlTree< Integer>( );
        final int NUMS = 200;
        final int GAP  =   17;
        System.out.println( "Checking... (no more output means success)" );
        for( int i = GAP; i != 0; i = ( i + GAP ) % NUMS ) {
            t.insert( i );
        }
        t.printTree( );
        System.out.println(t.height(t.root));
    }
}

完全二叉树

​ （1）若设二叉树的深度为k，除第 k 层外，其它各层 (1～k-1) 的结点数都达到最大个数，

​ （2）第k 层所有的结点都连续集中在最左边，这就是完全二叉树。

满二叉树

一棵二叉树的结点要么是叶子结点，要么它有两个子结点

​ 如果一个二叉树的层数为K，且结点总数是(2^k) -1，则它就是满二叉树

6.1.2 堆（来自数组的树）

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堆的性质，堆通常是一个可以被看做一棵树的数组对象。堆总是满足下列性质：
1.堆中某个节点的值总是不大于或不小于其父节点的值。
2.堆总是一棵完全二叉树。

public class Heap {
    private int[] arr;       // 堆是完全二叉树，底层用数组存储
    private int capacity;    // 堆中能存储的最大元素数量
    private int n;          // 当前堆中元素数量
    public Heap(int count) {
        capacity = count;
        arr = new int[capacity+1];
        n = 0;
    }
    public void insert(int value) {
        if (n >= capacity) {
            // 超过堆大小了，不能再插入元素
            return;
        }
        n++;
        // 先将元素插入到队尾中
        arr[n] = value;

        int i = n;
        // 由于我们构建的是一个大顶堆，所以需要不断调整以让其满足大顶堆的条件
        while (i/2 > 0 && arr[i] > arr[i/2]) {
            swap(arr, i, i/2);
            i = i / 2;
        }
    }
    /**
     * 移除堆顶元素
     */
    public void removeTopElement() {
        if (n == 0) {
            // 堆中如果没有元素，也就是不存在移除堆顶元素的情况了
            return;
        }
        int count = n;
        arr[1] = arr[count];
        --count;
        heapify(1, count);
    }
    /**
     * 自上而下堆化以满足大顶堆的条件
     */
    public void heapify(int index, int n) {

        while (true) {
            int maxValueIndex = index;
            if (2 * index <= n && arr[index] < arr[2 * index]) {
                // 左节点比其父节点大
                maxValueIndex = 2 * index;
            }

            if (2 * index + 1 <= n && arr[maxValueIndex] < arr[2 * index + 1]) {
                // 右节点比左节点或父节点大
                maxValueIndex = 2 * index + 1;
            }

            if (maxValueIndex == index) {
                // 说明当前节点值为最大值，无需再往下迭代了
                break;
            }
            swap(arr, index, maxValueIndex);
            index = maxValueIndex;
        }
    }
    /**
     * 交换数组第 i 和第 j 个元素
     */
    public static void swap(int[] arr, int i, int j)
    {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
}

第二种方法

import java.util.Arrays;

class Heap {
    private final static int N = 100; //default size
    private int[] nums;
    private int size;

    public Heap(int[] nums) {
        this.nums = nums;
        this.size = nums.length;
        heapify(this.nums);
    }
    public Heap() {
        this.nums = new int[N];
    }
    /**
     * heapify an array, O(n)
     * @param nums An array to be heapified.
     */
    private void heapify(int[] nums) {
        for (int j = (size - 1) >> 1; j >= 0; j--) {
            siftDown(j);
        }
    }
    /**
     * append x to heap
     * O(logn)
     * @param x
     * <a href='http://www.jobbole.com/members/wx1409399284'>@return</a>
     */
    public int insert(int x) {
        if (size >= this.nums.length) {
            expandSpace();
        }
        size += 1;
        nums[size-1] = x;
        siftUp(size-1);
        return x;
    }
    /**
     * delete an element located in i position.
     * O(logn)
     * @param i
     * <a ref='http://www.jobbole.com/members/wx1409399284'>@return</a>
     */
    public int delete(int i) {
        rangeCheck(i);
        int key = nums[i];
//将last元素覆盖过来,先siftDown; 再视情况考虑是否siftUp;
        int last = nums[i] = nums[size-1];
        size--;
        siftDown(i);
//check #i的node的键值是否确实发生改变,若发生改变，则ok,否则为确保堆性质，则需要siftUp;
        if (i < size && nums[i] == last) {
            siftUp(i);
        }
        return key;
    }
    /**
     * remove the root of heap, return it's value, and adjust heap to maintain the heap's property.
     * O(logn)
     * <a href='http://www.jobbole.com/members/wx1409399284'>@return</a>
     */
    public int extractMin() {
        rangeCheck(0);
        int key = nums[0], last = nums[size-1];
        nums[0] = last;
        size--;
        siftDown(0);
        return key;
    }
    /**
     * return an element's index, if not exists, return -1;
     * O(n)
     * @param x
     * <a href='http://www.jobbole.com/members/wx1409399284'>@return</a>
     */
    public int search(int x) {
        for (int i = 0; i < size; i++) {
            if (nums[i] == x) {
                return i;
            }
        }
        return -1;
    }
    /**
     * return but does not remove the root of heap.
     * O(1)
     * <a href='http://www.jobbole.com/members/wx1409399284'>@return</a>
     */
    public int peek() {
        rangeCheck(0);
        return nums[0];
    }
    private void siftUp(int i) {
        int key = nums[i];
        for (; i > 0;) {
            int p = (i - 1) >>> 1;
            if (nums[p] <= key) {
                break;
            }
            nums[i] = nums[p];
            i = p;
        }
        nums[i] = key;
    }
    private void siftDown(int i) {
        int key = nums[i];
        for (;i < size / 2;) {
            int child = (i << 1) + 1;
            if (child + 1 < size && nums[child] > nums[child+1]) {
                child++;
            }
            if (key <= nums[child]) {
                break;
            }
            nums[i] = nums[child];
            i = child;
        }
        nums[i] = key;
    }
    private void rangeCheck(int i) {
        if (!(0 <= i && i < size)) {
            throw new RuntimeException("Index is out of boundary");
        }
    }
    private void expandSpace() {
        this.nums = Arrays.copyOf(this.nums, size * 2);
    }
    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        sb.append("[");
        for (int i = 0; i < size; i++) {
            sb.append(String.format((i != 0 ? ", " : "") + "%d", nums[i]));
        }
        sb.append("] ");
        return sb.toString();
    }
}

堆的应用：海量实数中（一亿级别以上）找到TopK（一万级别以下）的数集合

import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.io.UnsupportedEncodingException;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;
public class TopKNumbersInMassiveNumbersDemo {
    public static void main(String[] args) {
// TODO Auto-generated method stub
        int[] topK = new int[]{50001,50002,50003,50004,50005};
        genData(1000 * 1000 * 1000, 500, topK);
        long t = System.currentTimeMillis();
        findTopK(topK.length);
        System.out.println(String.format("cost:%fs", (System.currentTimeMillis() - t) * 1.0 / 1000));
    }
    public static void genData(int N, int maxRandomNumer, int[] topK) {
        File f = new File("data.txt");
        int k = topK.length;
        Set<Integer> index = new TreeSet<>();
        for (;;) {
            index.add((int)(Math.random() * N));
            if (index.size() == k) {
                break;
            }
        }
        System.out.println(index);
        int j = 0;
        try {
            PrintWriter pW = new PrintWriter(f, "UTF-8");
            for (int i = 0; i < N; i++) {
                if(!index.contains(i)) {
                    pW.println((int)(Math.random() * maxRandomNumer));
                } else {
                    pW.println(topK[j++]);
                }
            }
            pW.flush();
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        }
    }
    public static void findTopK(int k) {
        int[] nums = new int[k];
        File f = new File("data.txt");
        try {
            Scanner scanner = new Scanner(f);
            for (int j = 0;j < k; j++) {
                nums[j] = scanner.nextInt();
            }
            heapify(nums);
            //core
            while (scanner.hasNextInt()) {
                int a = scanner.nextInt();
                if (a <= nums[0]) {
                    continue;
                } else {
                    nums[0] = a;
                    siftDown(0, k, nums);
                }

            }
            System.out.println(Arrays.toString(nums));
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }
    //O(n), minimal heap
    public static void heapify(int[] nums) {
        int size = nums.length;
        for (int j = (size - 1) >> 1; j >= 0; j--) {
            siftDown(j, size, nums);
        }
    }
    private static void siftDown(int i, int n, int[] nums) {
        int key = nums[i];
        for (;i < (n >>> 1);) {
            int child = (i << 1) + 1;
            if (child + 1 < n && nums[child] > nums[child+1]) {
                child++;
            }
            if (key <= nums[child]) {
                break;
            }
            nums[i] = nums[child];
            i = child;
        }
        nums[i] = key;
    }
}

实现优先队列

第一种：

import java.util.Arrays;

public class PriorityQueue2 {
    int[] arr = new int[10];
    int size;
    //交换
    public void swap(int[] arr, int i, int j) {
        int tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
    }
    //向上调整
    public void shiftUp(int[] arr, int sz, int child) {
        int parent = (child - 1) / 2;
        while (child > 0) {
            if (arr[child] < arr[parent]) {
                swap(arr, child, parent);
                child = parent;
                parent = (child - 1) / 2;
            } else {
                break;
            }
        }
    }
    //向下调整
    public void shiftDown(int[] arr, int sz, int parent) {
        int child = 2 * parent + 1;
        while (child < sz) {
            if (child + 1 < sz && arr[child + 1] < arr[child]) {
                child++;
            }
            if (arr[child] < arr[parent]) {
                swap(arr, child, parent);
                parent = child;
                child = 2 * parent + 1;
            } else {
                break;
            }
        }
    }
    //入队
    public void offer(int value) {
        if (size == arr.length) {
            arr = Arrays.copyOf(arr, arr.length * 2);
        }
        arr[size++] = value;    //尾插
        shiftUp(arr, size, size - 1);   //向上调整
    }
    //出队
    public int poll() {
        if (size > 0) {
            int peek = arr[0];
            swap(arr, 0, size - 1);
            size--;
            shiftDown(arr, size, 0);
            return peek;
        }
        return -1;
    }
    //取队顶元素
    public int peek() {
        return arr[0];
    }
    //判断是否为空
    public boolean isEmpty() {
        return size == 0;
    }
}

第二种：

public class PriorityQueue {
    private int[] data;
    private int size;

    public PriorityQueue(int size){
        data = new int[size];
        this.size = 0;
    }

    public void push(int toInsert) throws Exception{
        if(size == data.length) {
            throw new Exception("Queue is full!");
        }
        if(size == 0){
            data[0] = toInsert;
        }else{
            int i = size -1;
            for( ; i >= 0 ; i--){
                if(data[i] < toInsert){
                    data[i+1] = data[i];
                }else{
                    break;
                }
            }
            data[i+1] = toInsert;
        }
        size++;
    }

    public int pop() throws Exception{
        if(this.size == 0) {
            throw new Exception("Queue is empty!");
        }
        return data[--size];
    }

    public int peek() throws Exception{
        if(this.size == 0) {
            throw new Exception("Queue is empty!");
        }
        return data[size-1];
    }

    public int size(){
        return this.size;
    }

    public boolean isEmpty(){
        return (size == 0);
    }

    public boolean isFull(){
        return (size == data.length);
    }
}

6.1.3 红黑树

性质1：每个节点要么是黑色，要么是红色。

性质2：根节点是黑色。

性质3：每个叶子节点（NIL）是黑色。

性质4：每个红色节点的两个子节点一定都是黑色。 不能有两个红色节点相连。

性质5：任意一节点到每个叶子节点的路径都包含数量相同的黑结点。

​ 从性质5又可以推出：

​ 性质5.1：如果一个节点存在黑子节点，那么该结点肯定有两个子节点。不然走另一条路就会少一层黑色结点。

6.1.4 字典树（Trie）

实现

import java.util.Map;
import java.util.TreeMap;

public class Trie {
    private Node root;
    private int size;
    private static class Node {
        public boolean isWord;
        public Map<Character, Node> next;
        public Node() {
            next = new TreeMap<>();
        }
        public Node(boolean isWord) {
            this();
            this.isWord = isWord;
        }
    }
    public Trie() {
        root = new Node();
    }
    public int size() {
        return size;
    }
    public boolean isEmpty() {
        return size == 0;
    }
    /**
     * 插入操作
     *
     * @param word 单词
     */
    public void add(String word) {
        Node current = root;
        char[] cs = word.toCharArray();
        for (char c : cs) {
            Node next = current.next.get(c);
            if (next == null) {
                current.next.put(c, new Node());
            }
            current = current.next.get(c);
        }
        //如果当前的node已经是一个word，则不需要添加
        if (!current.isWord) {
            size++;
            current.isWord = true;
        }
    }
    /**
     * 是否包含某个单词
     *
     * @param word 单词
     * @return 存在返回true，反之false
     */
    public boolean contains(String word) {
        Node current = root;
        for (int i = 0; i < word.length(); i++) {
            char c = word.charAt(i);
            Node node = current.next.get(c);
            if (node == null) {
                return false;
            }
            current = node;
        }
        //如果只存在 panda这个词，查询 pan，虽然有这3个字母，但是并不存在该单词
        return current.isWord;
    }
    /**
     * Trie是否包含某个前缀
     *
     * @param prefix 前缀
     * @return
     */
    public boolean containsPrefix(String prefix) {
        Node current = root;
        for (int i = 0; i < prefix.length(); i++) {
            char c = prefix.charAt(i);
            Node node = current.next.get(c);
            if (node == null) {
                return false;
            }
            current = node;
        }
        return true;
    }

    /*
     * 1，如果单词是另一个单词的前缀，只需要把该word的最后一个节点的isWord的改成false
     * 2，如果单词的所有字母的都没有多个分支，删除整个单词
     * 3，如果单词的除了最后一个字母，其他的字母有多个分支，
     */
    /**
     * 删除操作
     *
     * @param word
     * @return
     */
    public boolean remove(String word) {
        Node multiChildNode = null;
        int multiChildNodeIndex = -1;
        Node current = root;
        for (int i = 0; i < word.length(); i++) {
            Node child = current.next.get(word.charAt(i));
            //如果Trie中没有这个单词
            if (child == null) {
                return false;
            }
            //当前节点的子节点大于1个
            if (child.next.size() > 1) {
                multiChildNodeIndex = i;
                multiChildNode = child;
            }
            current = child;
        }
        //如果单词后面还有子节点
        if (current.next.size() > 0) {
            if (current.isWord) {
                current.isWord = false;
                size--;
                return true;
            }
            //不存在该单词，该单词只是前缀
            return false;
        }
        //如果单词的所有字母的都没有多个分支，删除整个单词
        if (multiChildNodeIndex == -1) {
            root.next.remove(word.charAt(0));
            size--;
            return true;
        }
        //如果单词的除了最后一个字母，其他的字母有分支
        if (multiChildNodeIndex != word.length() - 1) {
            multiChildNode.next.remove(word.charAt(multiChildNodeIndex + 1));
            size--;
            return true;
        }
        return false;
    }
}

6.2 经典例题源码

方法一：递归

class Solution {
    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        } else {
            int leftHeight = maxDepth(root.left);
            int rightHeight = maxDepth(root.right);
            return Math.max(leftHeight, rightHeight) + 1;
        }
    }
}

方法二：广度优先搜索

class Solution {
    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        int ans = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            while (size > 0) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
                size--;
            }
            ans++;
        }
        return ans;
    }
}

方法一：深度优先搜索

class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }

        if (root.left == null && root.right == null) {
            return 1;
        }

        int min_depth = Integer.MAX_VALUE;
        if (root.left != null) {
            min_depth = Math.min(minDepth(root.left), min_depth);
        }
        if (root.right != null) {
            min_depth = Math.min(minDepth(root.right), min_depth);
        }

        return min_depth + 1;
    }
}

方法二：广度优先搜索

class Solution {
    class QueueNode {
        TreeNode node;
        int depth;

        public QueueNode(TreeNode node, int depth) {
            this.node = node;
            this.depth = depth;
        }
    }

    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }

        Queue<QueueNode> queue = new LinkedList<QueueNode>();
        queue.offer(new QueueNode(root, 1));
        while (!queue.isEmpty()) {
            QueueNode nodeDepth = queue.poll();
            TreeNode node = nodeDepth.node;
            int depth = nodeDepth.depth;
            if (node.left == null && node.right == null) {
                return depth;
            }
            if (node.left != null) {
                queue.offer(new QueueNode(node.left, depth + 1));
            }
            if (node.right != null) {
                queue.offer(new QueueNode(node.right, depth + 1));
            }
        }

        return 0;
    }
}

一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。

方法一：自顶向下的递归

class Solution {
    public boolean isBalanced(TreeNode root) {
        if (root == null) {
            return true;
        } else {
            return Math.abs(height(root.left) - height(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);
        }
    }

    public int height(TreeNode root) {
        if (root == null) {
            return 0;
        } else {
            return Math.max(height(root.left), height(root.right)) + 1;
        }
    }
}

方法二：自底向上的递归

class Solution {
    public boolean isBalanced(TreeNode root) {
        return height(root) >= 0;
    }

    public int height(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = height(root.left);
        int rightHeight = height(root.right);
        if (leftHeight == -1 || rightHeight == -1 || Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        } else {
            return Math.max(leftHeight, rightHeight) + 1;
        }
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode[] listOfDepth(TreeNode tree) {
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.offer(tree);

        List<ListNode> res = new ArrayList<>();
        ListNode dummy = new ListNode(0);
        while (!queue.isEmpty()) {
            int size = queue.size();
            ListNode curr = dummy;
            for (int i = 0; i < size; i++) {
                TreeNode treeNode = queue.poll();
                curr.next = new ListNode(treeNode.val);
                if (treeNode.left != null) {
                    queue.offer(treeNode.left);
                }
                if (treeNode.right != null) {
                    queue.offer(treeNode.right);
                }
                curr = curr.next;
            }
            res.add(dummy.next);
            dummy.next = null;
        }

        return res.toArray(new ListNode[] {});
    }
}

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        TreeNode left = invertTree(root.left);
        TreeNode right = invertTree(root.right);
        root.left = right;
        root.right = left;
        return root;
    }
}

方法一：非递归

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        //记录遍历到的每个节点的父节点。
        Map<TreeNode, TreeNode> parent = new HashMap<>();
        Queue<TreeNode> queue = new LinkedList<>();
        parent.put(root, null);//根节点没有父节点，所以为空
        queue.add(root);
        //直到两个节点都找到为止。
        while (!parent.containsKey(p) || !parent.containsKey(q)) {
            //队列是一边进一边出，这里poll方法是出队，
            TreeNode node = queue.poll();
            if (node.left != null) {
                //左子节点不为空，记录下他的父节点
                parent.put(node.left, node);
                //左子节点不为空，把它加入到队列中
                queue.add(node.left);
            }
            //右节点同上
            if (node.right != null) {
                parent.put(node.right, node);
                queue.add(node.right);
            }
        }
        Set<TreeNode> ancestors = new HashSet<>();
        //记录下p和他的祖先节点，从p节点开始一直到根节点。
        while (p != null) {
            ancestors.add(p);
            p = parent.get(p);
        }
        //查看p和他的祖先节点是否包含q节点，如果不包含再看是否包含q的父节点……
        while (!ancestors.contains(q))
            q = parent.get(q);
        return q;
    }

方法二：递归写法

    public TreeNode lowestCommonAncestor(TreeNode cur, TreeNode p, TreeNode q) {
        if (cur == null || cur == p || cur == q)
            return cur;
        TreeNode left = lowestCommonAncestor(cur.left, p, q);
        TreeNode right = lowestCommonAncestor(cur.right, p, q);
        //如果left为空，说明这两个节点在cur结点的右子树上，我们只需要返回右子树查找的结果即可
        if (left == null)
            return right;
        //同上
        if (right == null)
            return left;
        //如果left和right都不为空，说明这两个节点一个在cur的左子树上一个在cur的右子树上，
        //我们只需要返回cur结点即可。
        return cur;
    }

方法一：递归

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root.val < p.val && root.val < q.val)
            return lowestCommonAncestor(root.right, p, q);
        if(root.val > p.val && root.val > q.val)
            return lowestCommonAncestor(root.left, p, q);
        return root;
    }
}

方法二：迭代

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        while(root != null) {
            if(root.val < p.val && root.val < q.val) // p,q 都在 root 的右子树中
                root = root.right; // 遍历至右子节点
            else if(root.val > p.val && root.val > q.val) // p,q 都在 root 的左子树中
                root = root.left; // 遍历至左子节点
            else break;
        }
        return root;
    }
}
// 优化：若可保证 p.val < q.valp.val<q.val ，则在循环中可减少判断条件。
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(p.val > q.val) { // 保证 p.val < q.val
            TreeNode tmp = p;
            p = q;
            q = tmp;
        }
        while(root != null) {
            if(root.val < p.val) // p,q 都在 root 的右子树中
                root = root.right; // 遍历至右子节点
            else if(root.val > q.val) // p,q 都在 root 的左子树中
                root = root.left; // 遍历至左子节点
            else break;
        }
        return root;
    }
}

方法一：递归

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return check(root, root);
    }
    public boolean check(TreeNode p, TreeNode q) {
        if (p == null && q == null) {
            return true;
        }
        if (p == null || q == null) {
            return false;
        }
        return p.val == q.val && check(p.left, q.right) && check(p.right, q.left);
    }
}

方法二：迭代

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return check(root, root);
    }
    public boolean check(TreeNode u, TreeNode v) {
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(u);
        q.offer(v);
        while (!q.isEmpty()) {
            u = q.poll();
            v = q.poll();
            if (u == null && v == null) {
                continue;
            }
            if ((u == null || v == null) || (u.val != v.val)) {
                return false;
            }
            q.offer(u.left);
            q.offer(v.right);
            q.offer(u.right);
            q.offer(v.left);
        }
        return true;
    }
}

上下反转二叉树

class Solution {
    // 处理之后的根节点
     TreeNode head;
    public  TreeNode upsideDownBinaryTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        dfs(root);  // 对 root 进行上下翻转
        return head;
    }
    public  TreeNode dfs(TreeNode node) {
        if (node == null) {
            return null;
        }
        if (node.left == null && node.right == null) {
            if (head == null) {
                // 最左边的节点即为实际的根节点
                head = node;
            }
            return node;
        }
        TreeNode left = dfs(node.left);
        TreeNode right = dfs(node.right);
        if (left != null) {
            // 左孩子的左子树为当前的右节点
            left.left = right;
            // 左孩子的右子树为当前父节点
            left.right = node;
        }
        // 清空的当前父节点的左右子树
        node.right = null;
        node.left = null;
        return node;
    }
}

中序遍历

class Solution {
    List<Integer> answer = new ArrayList<Integer>();
    int base, count, maxCount;

    public int[] findMode(TreeNode root) {
        dfs(root);
        int[] mode = new int[answer.size()];
        for (int i = 0; i < answer.size(); ++i) {
            mode[i] = answer.get(i);
        }
        return mode;
    }

    public void dfs(TreeNode o) {
        if (o == null) {
            return;
        }
        dfs(o.left);
        update(o.val);
        dfs(o.right);
    }

    public void update(int x) {
        if (x == base) {
            ++count;
        } else {
            count = 1;
            base = x;
        }
        if (count == maxCount) {
            answer.add(base);
        }
        if (count > maxCount) {
            maxCount = count;
            answer.clear();
            answer.add(base);
        }
    }
}

class Solution {
    LinkedList<List<Integer>> res = new LinkedList<>();
    LinkedList<Integer> path = new LinkedList<>(); 
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        recur(root, sum);
        return res;
    }
    void recur(TreeNode root, int tar) {
        if(root == null) return;
        path.add(root.val);
        tar -= root.val;
        if(tar == 0 && root.left == null && root.right == null)
            res.add(new LinkedList(path));
        recur(root.left, tar);
        recur(root.right, tar);
        path.removeLast();
    }
}

方法一 前序遍历+排序

public class Problem02 {

    private List<Integer> ansList;

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }

        ansList.add(root.val);
        dfs(root.left);
        dfs(root.right);
    }

    public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
        ansList = new ArrayList<>();
        dfs(root1);
        dfs(root2);
        Collections.sort(ansList);
        return ansList;
    }

}

方法二 分别中序遍历+归并

public class Problem02_1 {

    private void dfs(TreeNode root, List<Integer> ansList) {
        if (root == null) {
            return;
        }

        dfs(root.left, ansList);
        ansList.add(root.val);
        dfs(root.right, ansList);
    }

    private List<Integer> merge(List<Integer> list1, List<Integer> list2) {
        List<Integer> ansList = new ArrayList<>();
        int size1 = list1.size();
        int size2 = list2.size();
        int index1, index2;
        for (index1 = 0, index2 = 0; index1 < size1 && index2 < size2;) {
            int num1 = list1.get(index1);
            int num2 = list2.get(index2);
            if (num1 < num2) {
                ansList.add(num1);
                index1++;
            } else {
                ansList.add(num2);
                index2++;
            }
        }

        while (index1 < size1) {
            ansList.add(list1.get(index1++));
        }

        while (index2 < size2) {
            ansList.add(list2.get(index2++));
        }

        return ansList;
    }

    public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
        List<Integer> ansList1 = new ArrayList<>();
        List<Integer> ansList2 = new ArrayList<>();
        dfs(root1, ansList1);
        dfs(root2, ansList2);

        return merge(ansList1, ansList2);
    }

}

方法三 遍历+优先队列

public class Problem02_2 {

    private PriorityQueue<Integer> priorityQueue;

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }

        priorityQueue.offer(root.val);
        dfs(root.left);
        dfs(root.right);
    }

    public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
        priorityQueue = new PriorityQueue<>();
        dfs(root1);
        dfs(root2);
        List<Integer> ansList = new ArrayList<>();
        while (!priorityQueue.isEmpty()) {
            ansList.add(priorityQueue.poll());
        }
        return ansList;
    }

}

方法一：深度优先搜索

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> paths = new ArrayList<String>();
        constructPaths(root, "", paths);
        return paths;
    }

    public void constructPaths(TreeNode root, String path, List<String> paths) {
        if (root != null) {
            StringBuffer pathSB = new StringBuffer(path);
            pathSB.append(Integer.toString(root.val));
            if (root.left == null && root.right == null) {  // 当前节点是叶子节点
                paths.add(pathSB.toString());  // 把路径加入到答案中
            } else {
                pathSB.append("->");  // 当前节点不是叶子节点，继续递归遍历
                constructPaths(root.left, pathSB.toString(), paths);
                constructPaths(root.right, pathSB.toString(), paths);
            }
        }
    }
}

方法二：广度优先搜索

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> paths = new ArrayList<String>();
        if (root == null) {
            return paths;
        }
        Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();
        Queue<String> pathQueue = new LinkedList<String>();

        nodeQueue.offer(root);
        pathQueue.offer(Integer.toString(root.val));

        while (!nodeQueue.isEmpty()) {
            TreeNode node = nodeQueue.poll(); 
            String path = pathQueue.poll();

            if (node.left == null && node.right == null) {
                paths.add(path);
            } else {
                if (node.left != null) {
                    nodeQueue.offer(node.left);
                    pathQueue.offer(new StringBuffer(path).append("->").append(node.left.val).toString());
                }

                if (node.right != null) {
                    nodeQueue.offer(node.right);
                    pathQueue.offer(new StringBuffer(path).append("->").append(node.right.val).toString());
                }
            }
        }
        return paths;
    }
}

class Solution {
    private boolean balanced = true;    // 默认是平衡的
    public boolean isBalanced(TreeNode root) {
        treeDepth(root);
        return balanced;
    }
    /*
        使用DFS计算以root为根的树的高度
        DFS的优点是最多访问所有节点一次
        先判断子树的平衡性，如果子树不平衡则说明整棵树是不平衡的，提前终止遍历
    */
    public int treeDepth(TreeNode root) {
        // 空树的高度为0
        // !balanced返回0完全是为了终止递归，其实返回什么无所谓
        if (root == null || !balanced) {    
            return 0;
        }
        int leftTreeDepth = treeDepth(root.left);       // 左子树高度
        int rightTreeDepth = treeDepth(root.right);     // 右子树高度
        // 平衡性判断
        if (Math.abs(leftTreeDepth - rightTreeDepth) > 1) {
            balanced = false;
        }
        // root树的高度应该是左右子树较大者 + 1
        return Math.max(leftTreeDepth, rightTreeDepth) + 1;
    }
}
//方法二：
class Solution {
    public boolean isBalanced(TreeNode root) {
        // 根结点为null，说明是棵空树，认为是平衡的
        if (root == null) {
            return true;
        }
        int leftTreeDepth = treeDepth(root.left);   // 左子树高度
        int rightTreeDepth = treeDepth(root.right); // 右子树高度
        if (Math.abs(leftTreeDepth - rightTreeDepth) > 1) {
            return false;   // 高度差大于1说明不平衡
        }
        // 若以当前节点为根的子树是平衡的，继续递归检查它左右子树的平衡性
        return isBalanced(root.left) && isBalanced(root.right); 
    }
    public int treeDepth(TreeNode root) {
        // 递归出口，空树的高度为0
        if (root == null) {
            return 0;
        }
        // 当前二叉树的高度 = max(左子树高度,右子树高度) + 1
        return Math.max(treeDepth(root.left), treeDepth(root.right)) + 1;
    }
}

方法一：宽度优先搜索

public List<Integer> largestValues(TreeNode root) {
    //LinkedList实现队列
    Queue<TreeNode> queue = new LinkedList<>();
    List<Integer> values = new ArrayList<>();
    if (root != null)
        queue.add(root);//入队
    while (!queue.isEmpty()) {
        int max = Integer.MIN_VALUE;
        int levelSize = queue.size();//每一层的数量
        for (int i = 0; i < levelSize; i++) {
            TreeNode node = queue.poll();//出队
            max = Math.max(max, node.val);//记录每层的最大值
            if (node.left != null)
                queue.add(node.left);
            if (node.right != null)
                queue.add(node.right);
        }
        values.add(max);
    }
    return values;
}

方法二：深度优先搜索

public List<Integer> largestValues(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    helper(root, res, 1);
    return res;
}
//level表示的是第几层，集合res中的第一个数据表示的是
// 第一层的最大值，第二个数据表示的是第二层的最大值……
private void helper(TreeNode root, List<Integer> res, int level) {
    if (root == null)
        return;
    //如果走到下一层了直接加入到集合中
    if (level == res.size() + 1) {
        res.add(root.val);
    } else {
        //注意：我们的level是从1开始的，也就是说root
        // 是第一层，而集合list的下标是从0开始的，
        // 所以这里level要减1。
        // Math.max(res.get(level - 1), root.val)表示的
        // 是遍历到的第level层的root.val值和集合中的第level
        // 个元素的值哪个大，就要哪个。
        res.set(level - 1, Math.max(res.get(level - 1), root.val));
    }
    //下面两行是DFS的核心代码
    helper(root.left, res, level + 1);
    helper(root.right, res, level + 1);
}

方法一：深度优先搜索

class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Integer> counts = new ArrayList<Integer>();
        List<Double> sums = new ArrayList<Double>();
        dfs(root, 0, counts, sums);
        List<Double> averages = new ArrayList<Double>();
        int size = sums.size();
        for (int i = 0; i < size; i++) {
            averages.add(sums.get(i) / counts.get(i));
        }
        return averages;
    }

    public void dfs(TreeNode root, int level, List<Integer> counts, List<Double> sums) {
        if (root == null) {
            return;
        }
        if (level < sums.size()) {
            sums.set(level, sums.get(level) + root.val);
            counts.set(level, counts.get(level) + 1);
        } else {
            sums.add(1.0 * root.val);
            counts.add(1);
        }
        dfs(root.left, level + 1, counts, sums);
        dfs(root.right, level + 1, counts, sums);
    }
}

方法二：广度优先搜索

class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> averages = new ArrayList<Double>();
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            double sum = 0;
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                sum += node.val;
                TreeNode left = node.left, right = node.right;
                if (left != null) {
                    queue.offer(left);
                }
                if (right != null) {
                    queue.offer(right);
                }
            }
            averages.add(sum / size);
        }
        return averages;
    }
}