55 - I. 二叉树的深度

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difficulty: 简单
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面试题 55 - I. 二叉树的深度

题目描述

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：

1. 节点总数 <= 10000

注意：本题与主站 104 题相同：https://leetcode.cn/problems/maximum-depth-of-binary-tree/

解法

方法一：递归

我们可以用递归的方法来解决这道题。递归的终止条件是当前节点为空，此时深度为 $0$；如果当前节点不为空，则当前的深度为其左右子树深度的最大值加 $1$，递归计算当前节点的左右子节点的深度，然后返回它们的最大值加 $1$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。最坏情况下，二叉树退化为链表，递归深度达到 $n$，系统使用 $O(n)$ 大小的栈空间。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def maxDepth(self, root: TreeNode) -> int:if root is None:return 0return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

Java

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

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;}return 1 + max(maxDepth(root->left), maxDepth(root->right));}
};

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}l, r := maxDepth(root.Left), maxDepth(root.Right)if l > r {return 1 + l}return 1 + r
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {pub fn max_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {match root {None => 0,Some(node) => {let mut node = node.borrow_mut();let left = node.left.take();let right = node.right.take();1 + Self::max_depth(left).max(Self::max_depth(right))}}}
}

JavaScript

/*** 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) {return 0;}return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
};

C#

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

Swift

/* public class TreeNode {
*     public var val: Int
*     public var left: TreeNode?
*     public var right: TreeNode?
*     public init(_ val: Int) {
*         self.val = val
*         self.left = nil
*         self.right = nil
*     }
* }
*/class Solution {func maxDepth(_ root: TreeNode?) -> Int {guard let root = root else {return 0}return 1 + max(maxDepth(root.left), maxDepth(root.right))}
}