# LeetCode – Maximum Depth of Binary Tree

### 题目：

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.

### 解题：

深度优先搜索方式，分别找到左右子树最大的深度+1即可；广度优先搜索，逐层遍历找到最大层数返回。

### 实现

• DFS， 空间复杂度O(N)    时间复杂度 O(N)

``````/**
* 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) {
if (!root) {return 0;}
return 1 + max( maxDepth(root->left), maxDepth(root->right));
}
};``````

### 实现：

• BFS  空间复杂度O(N)    时间复杂度 O(N)
``````/**
* 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) {
if (!root) {return 0;}
return 1 + max( maxDepth(root->left), maxDepth(root->right));
}
};*/

class Solution {
public:
int maxDepth(TreeNode* root) {
if (!root){
return 0;
}
queue<TreeNode*> q;
int depth=0;
q.push(root);
while(!q.empty()){
depth ++;
int level_size = q.size();
for (int i=0; i<level_size; i++){
TreeNode *node = q.front();
q.pop();
if (node->left){
q.push(node->left);
}
if (node->right){
q.push(node->right);
}
}
}
return depth;
}
};``````