LeetCode 솔루션

[Easy - wk4 - Q3] 104. Maximum Depth of Binary Tree

Chloe 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.05.23 00:53

6,954 조회
3 댓글
0 추천
0 비추천
목록

글검색

게시글 쓰기

Given the root of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 3

Example 2:

Input: root = [1,null,2]
Output: 2

Constraints:

The number of nodes in the tree is in the range [0, 104].
-100 <= Node.val <= 100

#leetcode, #문제풀이, #easy, #linkedin, #amazon, #spotify, #microsoft, #google, #bloomberg, #apple, #fasebook, #tree, #depth-first search, #breadth-first search, #binary tree

익명
작성일 2022.05.24 09:01

C++
Runtime: 12 ms, faster than 55.34% of C++ online submissions for Maximum Depth of Binary Tree.
Memory Usage: 18.9 MB, less than 14.43% of C++ online submissions for Maximum Depth of Binary Tree.

class Solution {
public:
    int maxDepth(TreeNode* root) {
        if (!root) return 0;
        return max(maxDepth(root->left), maxDepth(root->right)) + 1;
    }
};

익명
작성일 2022.06.01 20:43

JS
Runtime: 76 ms, faster than 78.73% of JavaScript online submissions for Maximum Depth of Binary Tree.
Memory Usage: 45 MB, less than 63.27% of JavaScript online submissions for Maximum Depth of Binary Tree.

var maxDepth = function(root) {
    let counter = 0
    let BFS = (root, level) => {
        if (!root) return;
        if (counter < level) {
            counter = level;
        }
        BFS(root.left, level+1);
        BFS(root.right,level+1);
    } 
    BFS(root, 1);
    return counter;
}

익명
작성일 2022.06.01 21:29

Java

class Solution {
  public int maxDepth(TreeNode root) {
    if (root == null) {
      return 0;
    } else {
      int left_height = maxDepth(root.left);
      int right_height = maxDepth(root.right);
      return java.lang.Math.max(left_height, right_height) + 1;
    }
  }
}