Check if max sum level of Binary tree divides tree into two equal sum halves

Given a Binary Tree, the task is to check if the maximum sum level divides the binary tree into the two parts of two equal sum halves.

Examples:

Input: 
                1 
              /   \ 
             2      3 
           /  \      \ 
          4    5      8 
                    /   \ 
                   2     4 
Output: YES
Explanation:
The maximum sum level is 2 and 
its sum is (4 + 5 + 8 = 17)
Sum of the upper half (1 + 2 + 3) = 6
Sum of the Lower half (2 + 4) = 6

Input:
                10 
              /    \ 
             20     30 
            /  \      \ 
           4    5      1 
Output: YES
Explanation:
The maximum sum level is 1 and 
its sum is (20 + 30 = 50)
Sum of the upper half (10) = 10
Sum of the lower half (5 + 4 + 1) = 10

Approach: The idea is to use level order traversal to compute the sum of every level of the binary tree. Then, find the maximum sum of the in all the levels. Finally check that the total sum of all the levels less than the maximum level sum is equal to the total sum of the levels of the greater than the maximum level sum.

Below is the implementation of the above approach:

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// C++ implementation to check if
// maximum level sum divides the
// Binary tree into two equal sum halves
  
#include <bits/stdc++.h>
using namespace std;
  
// Structure of the node
struct Node {
    int data;
    struct Node *left, *right;
};
  
// Utility function to
// create a new node
struct Node* newNode(int x)
{
    struct Node* temp = new Node;
    temp->data = x;
    temp->left = temp->right = NULL;
    return temp;
};
  
// Function to  check if
// maximum level sum divides the
// Binary tree into two equal sum halves
bool check_horizontal(struct Node* root)
{
    // Vector used to store the sum 
    // of all levels of the Binary Tree
    vector<int> sumLevel;
      
    // In index variable we store the 
    // level of the maximum level sum
    int index = -1, maxSum = 0, level = 0;
  
    queue<Node*> q;
    q.push(root);
  
    while (!q.empty()) {
          
        // Varible to store the 
        // current level sum.
        int sum = 0;
          
        // Size of the Queue
        int n = q.size();
          
        // Loop to iterate over the 
        // elements nodes of current level
        for (int i = 0; i < n; i++) {
              
            // Inserting the next level 
            // elements to the Queue
            Node* temp = q.front();
            sum += temp->data;
            if (temp->left != NULL)
                q.push(temp->left);
            if (temp->right != NULL)
                q.push(temp->right);
                  
            // Popping out the current 
            // level element from the Queue
            q.pop();
        }
          
        // Storing the current level
        // sum into the vector
        sumLevel.push_back(sum);
          
        // Level of maximum 
        // horizontal sum line
        if (sum > maxSum) {
            maxSum = sum;
            index = level;
        }
        level++;
    }
    // Find the left half and right 
    // half sum and check if they are equal
    int leftSum = 0, rightSum = 0;
    for (int i = 0; i < index; i++) {
        leftSum += sumLevel[i];
    }
  
    for (int i = index + 1; 
         i < sumLevel.size(); i++) {
        rightSum += sumLevel[i];
    }
    return (leftSum == rightSum);
}
  
// Driver Code
int main()
{
  
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->right = newNode(8);
    root->right->right->left = newNode(2);
    root->right->right->right = newNode(4);
  
    // Condition to check if the 
    // maxumum sum level divides 
    // it into two equal half
    if (check_horizontal(root))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}
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Output:
YES

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