# 面试题55 - I

输入一棵二叉树的根节点，求该树的深度。从根节点到叶节点依次经过的节点（含根、叶节点）形成树的一条路径，最长路径的长度为树的深度。

```
例如：

给定二叉树 [3,9,20,null,null,15,7]，

    3
   / \
  9  20
    /  \
   15   7

返回它的最大深度 3 。
```

## 提示：

* 节点总数 <= 10000

## 注意：本题与主站 104 题相同：[https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/](https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/)

## Solutions

1. **recursion**

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

1. **bfs with queue**

```cpp
/**
 * 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) {
        queue<TreeNode *> q;
        if (root) q.push(root);
        int depth = 0;

        while (!q.empty()) {
            depth++;
            int size = q.size();
            while (size--) {
                root = q.front(); q.pop();
                if (root->left)
                    q.push(root->left);
                if (root->right)
                    q.push(root->right);
            }
        }

        return depth;
    }
};
```

1. **traversal with stack**
2. simply record the maximum size of stack while traversing

```cpp
```
