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Given a binary tree, find its minimum depth.

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

**Note:** A leaf is a node with no children.

**Example 1:**

**Input:** root = [3,9,20,null,null,15,7]
**Output:** 2

**Example 2:**

**Input:** root = [2,null,3,null,4,null,5,null,6]
**Output:** 5

**Constraints:**

- The number of nodes in the tree is in the range
`[0, 10`

.^{5}] `-1000 <= Node.val <= 1000`

```
public class Solution {
public int minDepth(TreeNode root) {
if(root == null) return 0;
int left = minDepth(root.left);
int right = minDepth(root.right);
return (left == 0 || right == 0) ? left + right + 1: Math.min(left,right) + 1;
}
}
```

```
int minDepth(TreeNode* root) {
if (root == NULL) return 0;
queue<TreeNode*> Q;
Q.push(root);
int i = 0;
while (!Q.empty()) {
i++;
int k = Q.size();
for (int j=0; j<k; j++) {
TreeNode* rt = Q.front();
if (rt->left) Q.push(rt->left);
if (rt->right) Q.push(rt->right);
Q.pop();
if (rt->left==NULL && rt->right==NULL) return i;
}
}
return -1; //For the compiler thing. The code never runs here.
}
```

```
def minDepth(self, root):
if not root:
return 0
queue = collections.deque([(root, 1)])
while queue:
node, level = queue.popleft()
if node:
if not node.left and not node.right:
return level
else:
queue.append((node.left, level+1))
queue.append((node.right, level+1))
```

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