104.Maximum Depth of Binary Tree

题目描述

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.

我的解法

解题思路

可以用递归来实现深度优先的方法

实现代码

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/**
* 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 == NULL)
return 0;
int l = maxDepth(root->left) + 1;
int r = maxDepth(root->right) + 1;
return l > r ? l : r;
}
};

Runtime: 16 ms, faster than 21.06% of C++ online submissions for Maximum Depth of Binary Tree.
Memory Usage: 19 MB, less than 100.00% of C++ online submissions for Maximum Depth of Binary Tree.

堆栈实现DFS

实现代码

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/**
* 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 == NULL) return 0;
stack<pair<TreeNode*, int>> s;
int deep = 1;
s.push(pair<TreeNode*, int>(root, deep));
while(!s.empty()){
TreeNode* curNode = s.top().first;
int curDeep = s.top().second;
s.pop();
if (curNode){
deep = max(deep, curDeep);
s.push(pair<TreeNode*, int>(curNode -> left, curDeep+1));
s.push(pair<TreeNode*, int>(curNode -> right, curDeep+1));
}
}
return deep;
}
};

Runtime: 8 ms, faster than 87.69% of C++ online submissions for Maximum Depth of Binary Tree.
Memory Usage: 19.2 MB, less than 90.11% of C++ online submissions for Maximum Depth of Binary Tree.

知识总结