Difference between sum of even and odd valued nodes in a Binary Tree

Given a binary tree, the task is to find the absolute difference between the even valued and the odd valued nodes in a binary tree.
Examples: 
 

Input:
      5
    /   \
   2     6
 /  \     \  
1    4     8
    /     / \ 
   3     7   9
Output: 5
Explanation:
Sum of the odd value nodes is:
5 + 1 + 3 + 7 + 9 = 25 
Sum of the even value nodes is:
2 + 6 + 4 + 8 = 20 
Absolute difference = (25 – 20) = 5.

Input:
      4
    /   \
   1     4
 /  \     \  
7    2     6
Output: 8


Approach: 
Follow the steps below to solve the problem:

  • Traverse each node in the tree and check if the value at that node is odd or even.
  • Update oddSum and evenSum accordingly after visiting each node.
  • After complete traversal of the tree, print the absolute difference between oddSum and evenSum.

Below is the implementation of the above approach:

C++

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// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
int oddsum = 0;
int evensum = 0;
int ans = 0;
  
struct node {
    int data;
    struct node* left;
    struct node* right;
};
  
struct node* newnode(int data)
{
    node* temp = new node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
  
// Function calculate the sum of
// odd and even value node
void OddEvenDifference(struct node*
                           root)
{
    // If root is NULL
    if (root == NULL) {
        return;
    }
    else {
        // Check if current root
        // is odd or even
        if (root->data % 2 == 0) {
            evensum += root->data;
        }
        else {
            oddsum += root->data;
        }
        // Call on the left subtree
        OddEvenDifference(root->left);
  
        // Call on the right subtree
        OddEvenDifference(root->right);
    }
}
  
// Driver Code
int main()
{
    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);
  
    OddEvenDifference(root);
  
    cout << abs(oddsum - evensum)
         << endl;
}

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Java

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// Java implementation of
// the above approach
class GFG{
  
static int oddsum = 0;
static int evensum = 0;
static int ans = 0;
  
static class node
{
    int data;
    node left;
    node right;
};
  
static node newnode(int data)
{
    node temp = new node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
  
// Function calculate the sum of
// odd and even value node
static void OddEvenDifference(node root)
{
      
    // If root is null
    if (root == null
    {
        return;
    }
    else 
    {
          
        // Check if current root
        // is odd or even
        if (root.data % 2 == 0
        {
            evensum += root.data;
        }
        else 
        {
            oddsum += root.data;
        }
          
        // Call on the left subtree
        OddEvenDifference(root.left);
  
        // Call on the right subtree
        OddEvenDifference(root.right);
    }
}
  
// Driver Code
public static void main(String[] args)
{
    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);
  
    OddEvenDifference(root);
  
    System.out.print(Math.abs(
        oddsum - evensum) + "\n");
}
}
  
// This code is contributed by amal kumar choubey

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C#

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// C# implementation of
// the above approach
using System;
  
class GFG{
  
static int oddsum = 0;
static int evensum = 0;
//static int ans = 0;
  
class node
{
    public int data;
    public node left;
    public node right;
};
  
static node newnode(int data)
{
    node temp = new node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
  
// Function calculate the sum of
// odd and even value node
static void OddEvenDifference(node root)
{
      
    // If root is null
    if (root == null
    {
        return;
    }
    else
    {
          
        // Check if current root
        // is odd or even
        if (root.data % 2 == 0) 
        {
            evensum += root.data;
        }
        else
        {
            oddsum += root.data;
        }
          
        // Call on the left subtree
        OddEvenDifference(root.left);
  
        // Call on the right subtree
        OddEvenDifference(root.right);
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    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);
  
    OddEvenDifference(root);
  
    Console.Write(Math.Abs(
        oddsum - evensum) + "\n");
}
}
  
// This code is contributed by amal kumar choubey

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Output: 

5

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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