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++

// 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 

C#

// 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 

Output:

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

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My Personal Notes

C++

Java

C#

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