Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on.
Return the smallest level x such that the sum of all the values of nodes at level x is maximal.
Example 1:
Input: root = [1,7,0,7,-8,null,null] Output: 2 Explanation: 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.
Example 2:
Input: root = [989,null,10250,98693,-89388,null,null,null,-32127] Output: 2
Constraints:
[1, 104].-105 <= Node.val <= 105
class Solution {
public:
int bfs(TreeNode* root)
{
queue<TreeNode*> q;
int n,sum=0,level=1,maxm=INT_MIN,maxLevel;
q.push(root);
while(!q.empty()){
n = q.size();
sum = 0;
while(n--){
TreeNode* node = q.front();q.pop();
sum += node->val;
if(node->left)q.push(node->left);
if(node->right)q.push(node->right);
}
if(sum>maxm){
maxm = sum;
maxLevel = level;
}
level++;
}
return maxLevel;
}
int maxLevelSum(TreeNode* root) {
return bfs(root);
}
};
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