<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>

/**
 * 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 maxLevelSum(TreeNode* root) {
        queue<TreeNode*> curLvl;
        vector<TreeNode*> nextLvl;

        int curSum = 0;
        int lvl = 1;
        pair<int, int> ans = {1, root->val};

        curLvl.push(root);

        while(!curLvl.empty()){
            while(!curLvl.empty()){
                
                curSum += curLvl.front()->val;
                if(curLvl.front()->left)nextLvl.push_back(curLvl.front()->left);
                if(curLvl.front()->right)nextLvl.push_back(curLvl.front()->right);
                curLvl.pop();
            }

            for(auto node : nextLvl){
                curLvl.push(node);
            }

            nextLvl.clear();
            nextLvl.resize(0);
            if(curSum > ans.second){
                ans.second = curSum;
                ans.first = lvl;
            }
            curSum = 0;
            ++lvl;
        }
        return ans.first;
    }
};
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