# Maximum Level Sum of a Binary Tree in C++

C++Server Side ProgrammingProgramming

Suppose we have the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on. We have to return the smallest level X such that the sum of all the values of nodes at level X is maximal. So if the tree is like − Then the output will be 2, The level 1 sum = 1, level 2 sum is 7 + 0 = 7, level 2 sum is 7 + (-8) = -1, so max is of level 2, so output is 2.

To solve this, we will follow these steps −

• level := 1, sum := value of r, ansLevel := level, ansSum := sum
• define a queue q, insert given node r into q
• while q is not empty
• capacity := size of q
• increase level by 1, sum := 0
• while capacity is not 0
• node := front node from q, delete node from q
• if right of node is valid, then sum := sum + value of right node, insert right node into q
• if left of node is valid, then sum := sum + value of left node, insert left node into q
• decrease capacity by 1
• if ansSum < sum, then ansSum := sum, ansLevel := level
• return ansLevel

## Example

Let us see the following implementation to get better understanding −

class Solution {
public:
int maxLevelSum(TreeNode* r) {
int level = 1, sum = r->val;
int ansLevel = level, ansSum = sum;
queue <TreeNode*> q;
q.push(r);
while(!q.empty()){
int capacity = q.size();
level++;
sum = 0;
while(capacity--){
TreeNode* node = q.front();
q.pop();
if(node->right){
sum += node->right->val;
q.push(node->right);
}
if(node->left){
sum += node->left->val;
q.push(node->left);
}
}
if(ansSum<sum){
ansSum = sum;
ansLevel = level;
}
}
return ansLevel;
}
};

## Input

[1,7,0,7,-8,null,null]

## Output

2