comments	difficulty	edit_url
true	简单	https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9855%20-%20I.%20%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%B7%B1%E5%BA%A6/README.md

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 = None


class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if root is None:
            return 0
        return 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))
    }
}