<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> result;
        if (root == nullptr) return result;  // If root is null, return empty result
        
        queue<TreeNode*> q;
        q.push(root);
        
        while (!q.empty()) {
            int size = q.size();
            vector<int> currentLevel;  // Store all node values of the current level
            
            for (int i = 0; i < size; ++i) {
                TreeNode* node = q.front();
                q.pop();
                
                // Store the current node's value in the current level vector
                currentLevel.push_back(node->val);
                
                // Add children to the queue for the next level
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            
            // Find the max value in the current level and store it in result
            int maxVal = *max_element(currentLevel.begin(), currentLevel.end());
            result.push_back(maxVal);
        }
        
        return result;
    }
};

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