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

Last Updated : 21 Mar, 2023

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++

// 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;
}

Java

// 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

Python3

# Python3 implementation of
# the above approach
oddsum = 0
evensum = 0

class Node:
    
    def __init__(self, data):
        
        self.left = None
        self.right = None
        self.data = data
        
def newnode(data):

    temp = Node(data)
    
    return temp

# Function calculate the sum of
# odd and even value node
def OddEvenDifference(root):
    
    global evensum, oddsum
    
    # If root is NULL
    if (root == None):
        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
if __name__=="__main__":
    
    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)
    
    print(abs(oddsum - evensum))
    
# This code is contributed by rutvik_56

// 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

JavaScript

<script>
    // Javascript implementation of the above approach
    
    let oddsum = 0;
    let evensum = 0;
    let ans = 0;

    class node
    { 
        constructor(data) {
           this.data = data;
           this.left = this.right = null;
        }
    }
    
    function newnode(data)
    {
        let temp = new node(data);
        return temp;
    }

    // Function calculate the sum of
    // odd and even value node
    function OddEvenDifference(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);
        }
    }
    
    let 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);
 
    document.write(Math.abs(oddsum - evensum) + "</br>");

</script>

Output:

Time Complexity: O(N) where N is the number of nodes in given Binary Tree, as it does a simple traversal of the tree.
Auxiliary Space: O(h) where h is the height of given Binary Tree due to Recursion.

Another Approach (Iterative):
We can solve this problem by level order traversal using queue.

Below is the implementation of above approach:

C++

// C++ Program for the above approach
#include<bits/stdc++.h>
using namespace std;

// structure of node
struct Node{
    int data;
    struct Node* left = NULL;
    struct Node* right = NULL;
    Node(int data){
        this->data = data;
        this->left = this->right = NULL;
    }
};

// Utility function to create node
Node* newnode(int data){
    return new Node(data);
}

// function to find the absolute difference
int OddEvenDifference(Node* root){
    int oddSum = 0;
    int evenSum = 0;
    queue<Node*> q;
    q.push(root);
    while(!q.empty()){
        Node* front_node = q.front();
        q.pop();
        if(front_node->data % 2 == 0) evenSum += front_node->data;
        else oddSum += front_node->data;
        
        if(front_node->left != NULL) q.push(front_node->left);
        if(front_node->right != NULL) q.push(front_node->right);
    }
    return abs(oddSum - evenSum);
}

// 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);
 
    cout<<OddEvenDifference(root);
    return 0;
}
// This code is contributed by Kirti Agarwal(kirtiagarwal23121999)

Python3

# Python Program for the above approach
import queue

# structure of node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# function to find the absolute difference
def OddEvenDifference(root):
    oddSum = 0
    evenSum = 0
    # create a queue
    q = queue.Queue() 
    q.put(root) 
    while(not q.empty()): 
        front_node = q.get() # get the front node from the queue
        if(front_node.data % 2 == 0): # if the data of the front node is even
            evenSum += front_node.data # add it to the even sum
        else:
            oddSum += front_node.data # otherwise, add it to the odd sum
        
        if(front_node.left is not None): # if the front node has a left child
            q.put(front_node.left) # add it to the queue
        if(front_node.right is not None): # if the front node has a right child
            q.put(front_node.right) # add it to the queue
    return abs(oddSum - evenSum) # return the absolute difference between the odd and even sums

# Driver code
if __name__ == '__main__':
    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)
    
   # call the OddEvenDifference function and print the result
    print(OddEvenDifference(root))

Java

// Java Program for the above approach
import java.util.*;

// structure of node
class Node{
    int data;
    Node left;
    Node right;
    Node(int data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
};

// Driver code
class Main {
    // function to find the absolute difference
    static int OddEvenDifference(Node root){
        int oddSum = 0;
        int evenSum = 0;
        Queue<Node> q = new LinkedList<Node>();
        q.add(root);
        while(!q.isEmpty()){
            Node front_node = q.poll();
            if(front_node.data % 2 == 0) evenSum += front_node.data;
            else oddSum += front_node.data;

            if(front_node.left != null) q.add(front_node.left);
            if(front_node.right != null) q.add(front_node.right);
        }
        return Math.abs(oddSum - evenSum);
    }

    public static void main(String[] args) {
        Node root = new Node(5);
        root.left = new Node(2);
        root.right = new Node(6);
        root.left.left = new Node(1);
        root.left.right = new Node(4);
        root.left.right.left = new Node(3);
        root.right.right = new Node(8);
        root.right.right.right = new Node(9);
        root.right.right.left = new Node(7);

        System.out.println(OddEvenDifference(root));
    }
}

JavaScript

// JavaScript program for the above approach
// class of tree node
class Node{
    constructor(data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

// utility function to create node
function newnode(data){
    return new Node(data);
}

// function to find the absolute difference
function oddEvenDifference(root){
    let oddSum = 0;
    let evenSum = 0;
    let q = [];
    q.push(root);
    while(q.length > 0){
        let front_node = q.shift();
        if(front_node.data % 2 == 0) evenSum += front_node.data;
        else oddSum += front_node.data;
        
        if(front_node.left) q.push(front_node.left);
        if(front_node.right) q.push(front_node.right);
    }
    return Math.abs(oddSum - evenSum);
}

// driver code
let 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);

console.log(oddEvenDifference(root));

// C# Program for the above approach
using System;
using System.Collections.Generic;

// structure of node
public class Node {
    public int data;
    public Node left = null;
    public Node right = null;
    public Node(int data)
    {
        this.data = data;
        this.left = this.right = null;
    }
}

class MainClass {
    // Utility function to create node
    static Node newnode(int data) { return new Node(data); }
    // function to find the absolute difference
    static int OddEvenDifference(Node root)
    {
        int oddSum = 0;
        int evenSum = 0;
        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);
        while (q.Count > 0) {
            Node front_node = q.Peek();
            q.Dequeue();
            if (front_node.data % 2 == 0)
                evenSum += front_node.data;
            else
                oddSum += front_node.data;

            if (front_node.left != null)
                q.Enqueue(front_node.left);
            if (front_node.right != null)
                q.Enqueue(front_node.right);
        }
        return Math.Abs(oddSum - evenSum);
    }

    // Driver code
    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);

        Console.WriteLine(OddEvenDifference(root));
    }
}
// This code is contributed by user_dtewbxkn77n

Output

Difference between odd level and even level leaf sum in given Binary Tree

sunilkannur98

Improve

Article Tags :

Practice Tags :

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

Similar Reads

Thank You!

What kind of Experience do you want to share?