Given a a Binary Tree, find the difference between the sum of nodes at odd level and the sum of nodes at even level. Consider root as level 1, left and right children of root as level 2 and so on.

For example, in the following tree, sum of nodes at odd level is (5 + 1 + 4 + 8) which is 18. And sum of nodes at even level is (2 + 6 + 3 + 7 + 9) which is 27. The output for following tree should be 18 – 27 which is -9.

5 / \ 2 6 / \ \ 1 4 8 / / \ 3 7 9

A straightforward method is to **use level order traversal**. In the traversal, check level of current node, if it is odd, increment odd sum by data of current node, otherwise increment even sum. Finally return difference between odd sum and even sum. See following for implementation of this approach.

C implementation of level order traversal based approach to find the difference.

This approach is provided by Mandeep Singh. For **Iterative approach**, simply traverse the tree level by level (level order traversal), store sum of node values in even no. level in evenSum and rest in variable oddSum and finally return the difference.

Below is the simple implementation of the approach.

## C++

// CPP program to find // difference between // sums of odd level // and even level nodes // of binary tree #include <bits/stdc++.h> using namespace std; // tree node struct Node { int data; Node *left, *right; }; // returns a new // tree Node Node* newNode(int data) { Node* temp = new Node(); temp->data = data; temp->left = temp->right = NULL; return temp; } // return difference of // sums of odd level // and even level int evenOddLevelDifference(Node* root) { if (!root) return 0; // create a queue for // level order traversal queue<Node*> q; q.push(root); int level = 0; int evenSum = 0, oddSum = 0; // traverse until the // queue is empty while (!q.empty()) { int size = q.size(); level += 1; // traverse for // complete level while(size > 0) { Node* temp = q.front(); q.pop(); // check if level no. // is even or odd and // accordingly update // the evenSum or oddSum if(level % 2 == 0) evenSum += temp->data; else oddSum += temp->data; // check for left child if (temp->left) { q.push(temp->left); } // check for right child if (temp->right) { q.push(temp->right); } size -= 1; } } return (oddSum - evenSum); } // driver program int main() { // construct a tree Node* root = newNode(5); root->left = newNode(2); root->right = newNode(6); root->left->left = newNode(1); root->left->right = newNode(4); root->left->right->left = newNode(3); root->right->right = newNode(8); root->right->right->right = newNode(9); root->right->right->left = newNode(7); int result = evenOddLevelDifference(root); cout << "diffence between sums is :: "; cout << result << endl; return 0; } // This article is contributed by Mandeep Singh.

The problem can also be solved **using simple recursive traversal**. We can recursively calculate the required difference as, value of root’s data subtracted by the difference for subtree under left child and the difference for subtree under right child.

Below is the implementation of this approach.

## C

// A recursive program to find difference between sum of nodes at // odd level and sum at even level #include <stdio.h> #include <stdlib.h> // Binary Tree node struct node { int data; struct node* left, *right; }; // A utility function to allocate a new tree node with given data struct node* newNode(int data) { struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = node->right = NULL; return (node); } // The main function that return difference between odd and even level // nodes int getLevelDiff(struct node *root) { // Base case if (root == NULL) return 0; // Difference for root is root's data - difference for // left subtree - difference for right subtree return root->data - getLevelDiff(root->left) - getLevelDiff(root->right); } // Driver program to test above functions int main() { struct node *root = newNode(5); root->left = newNode(2); root->right = newNode(6); root->left->left = newNode(1); root->left->right = newNode(4); root->left->right->left = newNode(3); root->right->right = newNode(8); root->right->right->right = newNode(9); root->right->right->left = newNode(7); printf("%d is the required difference\n", getLevelDiff(root)); getchar(); return 0; }

## Java

// A recursive java program to find difference between sum of nodes at // odd level and sum at even level // A binary tree node class Node { int data; Node left, right; Node(int item) { data = item; left = right; } } class BinaryTree { // The main function that return difference between odd and even level // nodes Node root; int getLevelDiff(Node node) { // Base case if (node == null) return 0; // Difference for root is root's data - difference for // left subtree - difference for right subtree return node.data - getLevelDiff(node.left) - getLevelDiff(node.right); } // Driver program to test above functions public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(5); tree.root.left = new Node(2); tree.root.right = new Node(6); tree.root.left.left = new Node(1); tree.root.left.right = new Node(4); tree.root.left.right.left = new Node(3); tree.root.right.right = new Node(8); tree.root.right.right.right = new Node(9); tree.root.right.right.left = new Node(7); System.out.println(tree.getLevelDiff(tree.root) + " is the required difference"); } } // This code has been contributed by Mayank Jaiswal

## Python

# A recursive program to find difference between sum of nodes # at odd level and sum at even level # A Binary Tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # The main function that returns difference between odd and # even level nodes def getLevelDiff(root): # Base Case if root is None: return 0 # Difference for root is root's data - difference for # left subtree - difference for right subtree return (root.data - getLevelDiff(root.left)- getLevelDiff(root.right)) # Driver program to test above function root = Node(5) root.left = Node(2) root.right = Node(6) root.left.left = Node(1) root.left.right = Node(4) root.left.right.left = Node(3) root.right.right = Node(8) root.right.right.right = Node(9) root.right.right.left = Node(7) print "%d is the required difference" %(getLevelDiff(root)) # This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Output:

-9 is the required difference

Time complexity of both methods is O(n), but the second method is simple and easy to implement.

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This article is contributed by **Chandra Prakash**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.