<p>Given a binary tree, find its minimum depth.</p>

<p>The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.</p>

<p><strong>Note:</strong>&nbsp;A leaf is a node with no children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/12/ex_depth.jpg" style="width: 432px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [3,9,20,null,null,15,7]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [2,null,3,null,4,null,5,null,6]
<strong>Output:</strong> 5
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 10<sup>5</sup>]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>

# 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: Optional[TreeNode]) -> int:
        if not root:
            return 0
        def depth(root,lst=[],lev=1):
            if root and not root.left and not root.right:
                lst.append(lev)
                return lst
            if not root:
                return 
            depth(root.left,lst,lev+1)
            depth(root.right,lst,lev+1)
            return lst
        return min(depth(root))
42500