<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 level=1;
void levelOrderTraversal (TreeNode * root, unordered_map<int,int> &mp )  {

    queue <TreeNode*> q;
    q.push (root);
    q.push (NULL);

    while ( !q.empty() ) {
        TreeNode * temp=q.front();
        q.pop();
        if(temp!=NULL)
        mp[level]+=temp->val ;
        
        // temp=NULL means a level has been completely traversed and a newline character is needed
        if ( temp==NULL) {
            level++;
            // queue still has some child nodes present it means newline char is needed for that
            // particular level
            if ( !q.empty() ) 
             q.push(NULL);
        }
        
        else {

       if ( temp->left != NULL) 
        q.push( temp->left);
        
        if ( temp->right != NULL) 
            q.push(temp->right) ;
        
        }
    }
   
}
    int maxLevelSum(TreeNode* root) {
        // only 1 node
        if (root->left==NULL && root->right==NULL)
        return 1;

        unordered_map<int,int> mp; // (level,sum)
        levelOrderTraversal(root,mp);
        
        int maxsum=INT_MIN;
        int ans=0;
        for (auto it:mp) {
            if (it.second>maxsum) {
                maxsum=it.second;
                ans=it.first;
            }
            else if (it.second==maxsum && it.first<ans)
            ans=it.first;
        }

    return ans;
 
    }
};
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