题目描述

• 全局变量ans，遍历一遍树更新ans即可
• 带着 depth 跑 DFS
  

代码 & 思路

写成dfs了，确实是bfs

/*** 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 {// 全局变量maxint max = 0;public int maxDepth(TreeNode root) {dfs(root,0);return max;}// dfs遍历结点void dfs(TreeNode now, int depth){// 递归结束，进行max更新if(now == null){max = Math.max(depth, max);return;}dfs(now.left,depth+1);dfs(now.right,depth+1);}
}