<p>Given the <code>root</code> of a binary tree, return <em>an array of the largest value in each row</em> of the tree <strong>(0-indexed)</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/21/largest_e1.jpg" style="width: 300px; height: 172px;" />
<pre>
<strong>Input:</strong> root = [1,3,2,5,3,null,9]
<strong>Output:</strong> [1,3,9]
</pre>

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

<pre>
<strong>Input:</strong> root = [1,2,3]
<strong>Output:</strong> [1,3]
</pre>

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

<ul>
	<li>The number of nodes in the tree will be in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    vector<int> largestValues(TreeNode* root) {
        vector<int>v;
        vector<vector<int>>ans;
        queue<TreeNode*>q;
        q.push(root);
        if(q.front()==nullptr){
            return v;
        }
        while(1){
            int n=q.size();
            if(n==0){
                break;
            }
            vector<int>temp;
            while(n!=0){
                TreeNode* node=q.front();
                temp.push_back(node->val);
                q.pop();
                if(node->left!=nullptr){
                    q.push(node->left);
                }
                if(node->right!=nullptr){
                    q.push(node->right);
                }
                n--;
            }
            ans.push_back(temp);
        }
        for(int i=0;i<ans.size();i++){
            int a=*max_element(ans[i].begin(),ans[i].end());
            v.push_back(a);
        }
        return v;
    }
};
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