Question

104. 二叉树的最大深度

English Version

题目描述

给定一个二叉树 root ，返回其最大深度。

二叉树的 最大深度 是指从根节点到最远叶子节点的最长路径上的节点数。

 

示例 1：

 

输入：root = [3,9,20,null,null,15,7]
输出：3

示例 2：

输入：root = [1,null,2]
输出：2

 

提示：

• 树中节点的数量在 [0, 104] 区间内。
• -100 <= Node.val <= 100

解法

方法一：递归

递归遍历左右子树，求左右子树的最大深度，然后取最大值加 $1$ 即可。

时间复杂度 $O(n)$，其中 $n$ 是二叉树的节点数。每个节点在递归中只被遍历一次。

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

/**
 * 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;
    }
    int l = maxDepth(root.left);
    int r = maxDepth(root.right);
    return 1 + Math.max(l, r);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  int maxDepth(TreeNode* root) {
    if (!root) return 0;
    int l = maxDepth(root->left), r = maxDepth(root->right);
    return 1 + max(l, r);
  }
};

/**
 * 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)
  return 1 + max(l, r)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function maxDepth(root: TreeNode | null): number {
  if (root === null) {
    return 0;
  }
  return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
}

// 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::rc::Rc;
use std::cell::RefCell;
impl Solution {
  fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
    if root.is_none() {
      return 0;
    }
    let node = root.as_ref().unwrap().borrow();
    1 + Self::dfs(&node.left).max(Self::dfs(&node.right))
  }

  pub fn max_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
    Self::dfs(&root)
  }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxDepth = function (root) {
  if (!root) return 0;
  const l = maxDepth(root.left);
  const r = maxDepth(root.right);
  return 1 + Math.max(l, r);
};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   struct TreeNode *left;
 *   struct TreeNode *right;
 * };
 */

#define max(a, b) (((a) > (b)) ? (a) : (b))

int maxDepth(struct TreeNode* root) {
  if (!root) {
    return 0;
  }
  int left = maxDepth(root->left);
  int right = maxDepth(root->right);
  return 1 + max(left, right);
}