DESC:

给定一个二叉树，找出其最大深度。

二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。

说明: 叶子节点是指没有子节点的节点。

示例：
给定二叉树 [3,9,20,null,null,15,7]，

    3
   / \
9 20
    / \
   15   7

返回它的最大深度 3 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-depth-of-binary-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

CODE1:

JAVA:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode() {} *     TreeNode(int val) { this.val = val; } *     TreeNode(int val, TreeNode left, TreeNode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */class Solution {    public int maxDepth(TreeNode root) {        if (root == null) {            return 0;        }        return 1+Math.max(this.maxDepth(root.left), this.maxDepth(root.right));    }}

NOTES1:

1.递归，dfs

CODE2:

JAVA:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode() {} *     TreeNode(int val) { this.val = val; } *     TreeNode(int val, TreeNode left, TreeNode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */class Solution {    public int maxDepth(TreeNode root) {        if (root == null) {            return 0;        }        Queue
    
      queue = new LinkedList<>();        queue.offer(root);        int res = 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;            }            ++res;        }        return res;    }}