Difference between sums of odd position and even position nodes for each level of a Binary Tree

Given a Binary Tree, the task is to find the absolute difference between the sums of odd and even positioned nodes. A node is said to be odd and even positioned if its position in the current level is odd and even respectively. Note that the first element of each row is considered as odd positioned.

Examples:

Input:
      5
    /   \
   2     6
 /  \     \  
1    4     8
    /     / \ 
   3     7   9
Output: 11
Level     oddPositionNodeSum  evenPositionNodeSum
0             5                  0
1             2                  6
2             9                  4
3             12                 7
Difference = |(5 + 2 + 9 + 12) - (0 + 6 + 4 + 7)| = |28 - 17| = 11

Input:
      5
    /   \
   2     3
Output: 4

Approach: To find the sum of nodes at even and odd positions level by level, use level order traversal. While traversing the tree level by level mark flag oddPosition as true for the first element of each row and switch it for each next element of the same row. And in case if oddPosition flag is true add the node data into oddPositionNodeSum else add node data to evenPositionNodeSum. After completion of tree traversal find the absolute value of their differences at the end of the tree traversal.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
struct Node {
    int data;
    Node *left, *right;
};
  
// Iterative method to perform level
// order traversal line by line
int nodeSumDiff(Node* root)
{
  
    // Base Case
    if (root == NULL)
        return 0;
  
    int evenPositionNodeSum = 0;
    int oddPositionNodeSum = 0;
  
    // Create an empty queue for level
    // order traversal
    queue<Node*> q;
  
    // Enqueue root element
    q.push(root);
  
    while (1) {
  
        // nodeCount (queue size) indicates
        // number of nodes in the current level
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
  
        // Mark 1st node as even positioned
        bool oddPosition = true;
  
        // Dequeue all the nodes of current level
        // and Enqueue all the nodes of next level
        while (nodeCount > 0) {
            Node* node = q.front();
  
            // Depending upon node position
            // add value to their respective sum
            if (oddPosition)
                oddPositionNodeSum += node->data;
            else
                evenPositionNodeSum += node->data;
  
            q.pop();
            if (node->left != NULL)
                q.push(node->left);
            if (node->right != NULL)
                q.push(node->right);
            nodeCount--;
  
            // Switch the even position flag
            oddPosition = !oddPosition;
        }
    }
  
    // Return the absolute difference
    return abs(oddPositionNodeSum - evenPositionNodeSum);
}
  
// Utility method to create a node
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
  
// 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->left = newNode(6);
    root->right->right = newNode(7);
    root->left->right->left = newNode(8);
    root->left->right->right = newNode(9);
    root->left->right->right->right = newNode(10);
  
    cout << nodeSumDiff(root);
  
    return 0;
}

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Output:

7


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