## [Leetcode]104. Maximum Depth of Binary Tree(C++)

### 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 nearest leaf node.

### 例子

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

### Note

• A leaf is a node with no children.

### 解题思路

111. Minimum Depth of Binary Tree 可以用 BFS 或者 DFS 求解，代码如下：

#### 方法一： BFS

``````/**
* 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) {
std::queue<TreeNode*> q;
if (!root) return 0;
q.push(root);
int depth = 0;
while (!q.empty()) {
depth++;
std::queue<TreeNode*> next_level;
while (!q.empty()) {
TreeNode* node = q.front();
q.pop();
if (node->left) next_level.push(node->left);
if (node->right) next_level.push(node->right);
}
std::swap(q, next_level);
}

return depth;
}
};
``````
• 时间复杂度: O(n)
• 空间复杂度: O(w)

#### 方法二： DFS

``````class Solution {
public:
int maxDepth(TreeNode* root) {
if (!root) return 0;
int max_depth = 0;
dfs(root, 0, max_depth);

return max_depth;
}

private:
void dfs(TreeNode* node, int current_depth, int& max_depth) {
if (!node) {
return;
}
current_depth++;
if (!node->left && !node->right) {
max_depth = std::max(current_depth, max_depth);
}

dfs(node->left, current_depth, max_depth);
dfs(node->right, current_depth, max_depth);
}
};
``````
• 时间复杂度: O(n)
• 空间复杂度: O(h)

GitHub 代码同步地址： 104.MaximumDepthOfBinaryTree.cpp

Built with Hugo
Theme Stack designed by Jimmy