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[Easy - wk4 - Q3] 104. Maximum Depth of Binary Tree
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104. Maximum Depth of Binary Tree
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
관련자료
댓글 3
mingki님의 댓글
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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.
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;
}
};
dawn27님의 댓글
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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.
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;
}
Jack님의 댓글
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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;
}
}
}