# How to solve Leetcode 104. Maximum Depth of Binary Tree

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## Problem statement

Given the `root` of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

### Example 1

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

### Example 2

``````Input: root = [1,null,2]
Output: 2
``````

### Constraints

• The number of nodes in the tree is in the range `[0, 10^4]`.

• `-100 <= Node.val <= 100`.

## Solution

You have the following recursive relationship between the `root` and its children.

``````maxDepth(root) = max(maxDepth(root->left), maxDepth(root->right))
``````

### Code

``````#include <iostream>
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) {}
};

int maxDepth(TreeNode* root) {
if (root == nullptr) {
return 0;
}
return 1 + std::max(maxDepth(root->left), maxDepth(root->right));
}

int main() {
TreeNode fifteen(15);
TreeNode seven(7);
TreeNode twenty(20, &fifteen, &seven);
TreeNode nine(9);
TreeNode three(3, &nine, &twenty);
std::cout << maxDepth(&three) << std::endl;
TreeNode two(2);
TreeNode one(1, nullptr, &two);
std::cout << maxDepth(&one) << std::endl;
}
``````
``````Output:
3
2
``````

### Complexity

• Runtime: `O(N)`, where `N` is the number of nodes.

• Extra space: `O(N)`.

## References

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