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方法一 递归

/** * 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;        }        //如果为空  直接返回0        int a = maxDepth(root.left);        int b = maxDepth(root.right);        return Math.max(a, b) + 1;//加上根节点的层数            }}

方法二 广度优先搜索

/** * 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 ans = 0;        while(!queue.isEmpty()) {
               //记录每一行的节点个数 一行全部从队列拿完ans就加一            int size = queue.size();            while(size > 0) {
                   TreeNode cur = queue.poll();                //拿出一个size--                size--;                if(cur.left != null) {
                       queue.offer(cur.left);                }                if(cur.right != null) {
                       queue.offer(cur.right);                }            }            //一行拿完            ans++;        }        return ans;    }}
   

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