本问题对应的 leetcode 原文链接:剑指 Offer 55 – I. 二叉树的深度
问题描述
输入一棵二叉树的根节点,求该树的深度。从根节点到叶节点依次经过的节点(含根、叶节点)形成树的一条路径,最长路径的长度为树的深度。
例如:
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 。
提示:
节点总数 <= 10000
解题思路
视频讲解直达: 本题视频讲解
代码实现
class Solution {
public int maxDepth(TreeNode root) {
if(root == null) return 0;
int left = maxDepth(root.left);
int right = maxDepth(root.right);
return Math.max(left, right) + 1;
}
}
Python
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def maxDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
left = self.maxDepth(root.left)
right = self.maxDepth(root.right)
return max(left, right) + 1
C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxDepth(TreeNode* root) {
if (!root) {
return 0;
}
int left = maxDepth(root->left);
int right = maxDepth(root->right);
return max(left, right) + 1;
}
};
Go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func maxDepth(root *TreeNode) int {
if root == nil {
return 0
}
left := maxDepth(root.Left)
right := maxDepth(root.Right)
return max(left, right) + 1
}
func max(x, y int) int {
if x > y {
return x
}
return y
}
JS
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var maxDepth = function(root) {
if (root == null) return 0;
let left = maxDepth(root.left);
let right = maxDepth(root.right);
return Math.max(left, right) + 1;
};
评论(2)
思路:递归
● 时间复杂度:O(n)
● 空间复杂度: O(n)
时间复杂度o(n)
空间复杂度o(n)