<p>Given the <code>root</code> of a binary tree, the level of its root is <code>1</code>, the level of its children is <code>2</code>, and so on.</p>

<p>Return the <strong>smallest</strong> level <code>x</code> such that the sum of all the values of nodes at level <code>x</code> is <strong>maximal</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/03/capture.JPG" style="width: 200px; height: 175px;" />
<pre>
<strong>Input:</strong> root = [1,7,0,7,-8,null,null]
<strong>Output:</strong> 2
<strong>Explanation: </strong>
Level 1 sum = 1.
Level 2 sum = 7 + 0 = 7.
Level 3 sum = 7 + -8 = -1.
So we return the level with the maximum sum which is level 2.
</pre>

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

<pre>
<strong>Input:</strong> root = [989,null,10250,98693,-89388,null,null,null,-32127]
<strong>Output:</strong> 2
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    int maxLevelSum(TreeNode* root) {
        if(root==NULL) return {};

        queue<TreeNode*> que;
        que.push(root);

        int level = 1;
        int maxSum = INT_MIN;
        int maxLevel = 0;

        while(!que.empty()){
            int n = que.size();
            int sum = 0;
            TreeNode* node = NULL;

            while(n--){
                node = que.front();
                que.pop();
                sum += node->val;

                if(node->left) que.push(node->left);
                if(node->right) que.push(node->right);
            }

            if(sum > maxSum){
                maxSum = sum;
                maxLevel = level;
            }

            level++;
        }
        return maxLevel;
    }
};
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