#### Title description:

Enter the root node of a binary tree and find the depth of the tree. From the root node to the leaf node, the nodes (including the root and leaf nodes) passing through in turn form a path of the tree, and the length of the longest path is the depth of the tree.

For example, enter:

Given a binary tree [3,9,20,null,null,15,7],

```
3
/ \
9 20
/ \
15 7
```

**Return its maximum depth 3.**

#### data range:

```
节点总数 <= 10000
```

#### Problem-solving ideas:

- dfs, depth first search.
- The search order is the left subtree first, then the right subtree. Finally, return to this node, take the maximum depth of the left and right subtrees and +1
- The recursive exit is a non-leaf node (NULL), just return.

#### AC code (c++)

```
/**
* 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 dfs(TreeNode * root){
if (root == NULL){
return 0;
}
return max(dfs(root->left),dfs(root->right))+1;
}
int maxDepth(TreeNode* root) {
return dfs(root);
}
};
```