# 二叉树的最小深度

<p>给定一个二叉树，找出其最小深度。</p>

<p>最小深度是从根节点到最近叶子节点的最短路径上的节点数量。</p>

<p><strong>说明：</strong>叶子节点是指没有子节点的节点。</p>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/6d8500c55fe3ebe98dc781d7b62a3bbf.jpeg" style="width: 432px; height: 302px;" />
<pre>
<strong>输入：</strong>root = [3,9,20,null,null,15,7]
<strong>输出：</strong>2
</pre>

<p><strong>示例 2：</strong></p>

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

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>树中节点数的范围在 <code>[0, 10<sup>5</sup>]</code> 内</li>
	<li><code>-1000 <= Node.val <= 1000</code></li>
</ul>


## template

```python
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution:
    def minDepth(self, root: TreeNode) -> int:
        if not root:
            return 0
        queue = [root]
        count = 1
        while queue:

            next_queue = []
            for node in queue:
                if not node.left and not node.right:
                    return count
                if node.left:
                    next_queue.append(node.left)
                if node.right:
                    next_queue.append(node.right)
            queue = next_queue
            count += 1

        return count

```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```