<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 { // using level-order traversals -: 
public:
    
    int maxsum=INT_MIN;
    int anslevel=0;
    
    int levels(TreeNode* root){
        if(root==NULL) return 0;
        return 1+max(levels(root->left),levels(root->right));
    }
    
    void nthlevel(TreeNode* root, int curr, int level, int &currsum){
        if(root==NULL) return;
        
        if(curr==level){
            currsum+=root->val;
        }
        
        nthlevel(root->left, curr+1, level, currsum);
        nthlevel(root->right, curr+1, level, currsum);
    }
    
    // Level-order traversal to compute the sum at each level -:
    void levelorder(TreeNode* root){
        int n=levels(root);
        for(int i=1; i<=n; i++){
            int currsum=0;
            
            nthlevel(root,1,i,currsum); 
            
            // Update maximum sum and answer level
            if(currsum>maxsum){
                maxsum=currsum;
                anslevel=i;
            }
        }
    }
    
    int maxLevelSum(TreeNode* root) {
        levelorder(root);
        return anslevel;
    }
};
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