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# 111. Minimum Depth of Binary Tree

\# Easy

{% hint style="info" %}
This must be recursive problem, I sovled it in DFS.

The key is to think about the edge case and regular case.
{% endhint %}

![](https://3288217904-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LxJcc9A1TOyn5a5HJQ4%2F-MANHjFuN98xnjjsHTf5%2F-MANLeZNxZfw2yBbVx5N%2F1592766477380.jpg?alt=media\&token=0839d9b7-f1fd-41d3-9155-75c4dbf29807)

{% tabs %}
{% tab title="Python" %}

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def minDepth(self, root: TreeNode) -> int:
        # edge case
        if root == None:
            return 0
        elif root.left == None and root.right == None:
            return 1
        
        # regular case
        if root.left != None and root.right != None:
            min_left = self.minDepth(root.left)
            min_right = self.minDepth(root.right)
            minDepth = min(min_left, min_right) + 1
        elif root.left != None:
            minDepth = self.minDepth(root.left) + 1
        elif root.right != None:
            minDepth = self.minDepth(root.right) + 1
        
        return minDepth
```

{% endtab %}

{% tab title="Second Tab" %}

{% endtab %}
{% endtabs %}

Time complexity = $$O(n)$$ , space complexity = $$O(1)$$ . Because every node is traversalled and ervery node only stores its minimun depth.