題目描述

思路

解法1

/**
 public class TreeNode {
 int val = 0;
 TreeNode left = null;
 TreeNode right = null;

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

 }

 }
 */
public class Solution {
    public
 
 int TreeDepth(TreeNode root) {
        if (root == null)
            return 0;
        // 分別得到root左右孩子的深度
        int left = TreeDepth(root.left);
        int right = TreeDepth(root.right);
        return Math.max(left,right) + 1;
    }
}

解法2

import java.util.*;
//BFS的層次遍歷思想，記錄二叉樹的層數，
//遍歷完，層數即為最大深度
public
 
 class Solution {
    public int TreeDepth(TreeNode root) {
        if(root == null) return 0;
        int depth = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            int levelNums = queue.size();//先獲取本層的節點數，然後遍歷
            depth++;
            
 
for(int i = 0; i < levelNums; i++){
                TreeNode cur = queue.poll();
                if(cur.left != null) queue.offer(cur.left);
                if(cur.right != null) queue.offer(cur.right);
            }
        }
        return depth;
    }
}