<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.
 * 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) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if(!root)
            return 0;
        
        queue<TreeNode*> q;
        q.push(root);
        int depth=1;

        while(!q.empty())
        {
            int cnt=q.size();
            for (int i=0;i<cnt;i++)
            {
                TreeNode* node=q.front();
                q.pop();

                if(node->left==NULL && node->right==NULL )
                    return depth;
                if(node->left)
                    q.push(node->left);
                if(node->right)
                    q.push(node->right);
                
            }
            depth++;
        }
        return depth;
    }
};
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