104.Maximum Depth of Binary Tree¶
Tags: Easy Tree Depth-first Search
Links: https://leetcode.com/problems/maximum-depth-of-binary-tree/
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Note: A leaf is a node with no children.
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its depth = 3.
/**
* 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) {
std::ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
if (!root) return 0;
return 1 + max(maxDepth(root -> left), maxDepth(root -> right));
}
};
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Maximum Depth of Binary Tree.
Memory Usage: 18.4 MB, less than 100.00% of C++ online submissions for Maximum Depth of Binary Tree.
递归的方法。显然也会想到迭代的方法。
/**
* 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) {
std::ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
if (!root) return 0;
int res = 0;
queue<TreeNode *> q;
q.push(root);
while (!q.empty()) {
int len = q.size();
++res;
for (int i = 0; i < len; ++i) {
TreeNode *tmp = q.front(); q.pop();
if (tmp -> left) q.push(tmp -> left);
if (tmp -> right) q.push(tmp -> right);
}
}
return res;
}
};
Runtime: 4 ms, faster than 98.68% of C++ online submissions for Maximum Depth of Binary Tree.
Memory Usage: 18.4 MB, less than 100.00% of C++ online submissions for Maximum Depth of Binary Tree.
利用层序遍历的方法。