Count all pairs of adjacent nodes whose XOR is an odd number

Given a Binary Tree as shown below. The task is to count all pair of adjacent nodes whose XOR is an odd number.

Explanation:



Initially, root will be 0, start traversing the tree.
XOR of 15 and 13 will be  2 (Even)
XOR of 13 and 12 will be 1 (Odd)
XOR of 13 and 14 will be 5 (Even)
XOR of 15 and 18 will be 13 (Odd)
XOR of 18 and 17 will be 3 (Odd)
XOR of 18 and 21 will be 7 (Odd)

Therefore, total adjacent pairs with odd XOR = 5

Approach:

  1. Start traversing the tree from top to down.
  2. Every time perform XOR operation with the current node data and its adjacent data.
  3. If XOR of both node is an odd number then increment the count.

Below is the implementation of the above approach:

C++

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// CPP program to find number of adjacent pair
// in Binary Tree with odd xor
  
#include <iostream>
using namespace std;
  
// Tree Node
struct Node {
    int data;
    struct Node *left, *right;
};
  
// Function to find number of adjacent pair
// in Binary Tree with odd xor
int countOddXor(Node* root, Node *parent=NULL)
{
    // If Node is empty
    if (root == NULL)
        return 0;
  
    // check pair of XOR is odd or not
    int res = 0;
    if (parent != NULL && (parent->data ^ root->data) % 2) 
        res++;
  
    return res + countOddXor(root->left, root) + 
                 countOddXor(root->right, root);
}
  
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}
  
// Driver code
int main()
{
    struct Node* root = NULL;
  
    root = newNode(15);
    root->left = newNode(13);
    root->left->left = newNode(12);
    root->left->right = newNode(14);
    root->right = newNode(18);
    root->right->left = newNode(17);
    root->right->right = newNode(21);
  
    printf("%d ", countOddXor(root));
  
    return 0;
}

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Java

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// Java program to find number of adjacent pair
// in Binary Tree with odd xor
class GFG
{
  
// Tree Node
static class Node
{
    int data;
    Node left, right;
};
  
// Function to find number of adjacent pair
// in Binary Tree with odd xor
static int countOddXor(Node root, Node parent)
{
    // If Node is empty
    if (root == null)
        return 0;
  
    // check pair of XOR is odd or not
    int res = 0;
    if (parent != null && 
       (parent.data ^ root.data) % 2 == 1
        res++;
  
    return res + countOddXor(root.left, root) + 
                 countOddXor(root.right, root);
}
  
// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}
  
// Driver code
public static void main(String[] args)
{
    Node root = null;
  
    root = newNode(15);
    root.left = newNode(13);
    root.left.left = newNode(12);
    root.left.right = newNode(14);
    root.right = newNode(18);
    root.right.left = newNode(17);
    root.right.right = newNode(21);
  
    System.out.printf("%d ", countOddXor(root, null));
}
}
  
// This code is contributed by PrinciRaj1992

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C#

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// C# program to find number of adjacent pair
// in Binary Tree with odd xor
using System;
  
class GFG
{
  
// Tree Node
public class Node
{
    public int data;
    public Node left, right;
};
  
// Function to find number of adjacent pair
// in Binary Tree with odd xor
static int countOddXor(Node root, 
                       Node parent)
{
    // If Node is empty
    if (root == null)
        return 0;
  
    // check pair of XOR is odd or not
    int res = 0;
    if (parent != null && 
       (parent.data ^ root.data) % 2 == 1) 
        res++;
  
    return res + countOddXor(root.left, root) + 
                 countOddXor(root.right, root);
}
  
// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}
  
// Driver code
public static void Main(String[] args)
{
    Node root = null;
  
    root = newNode(15);
    root.left = newNode(13);
    root.left.left = newNode(12);
    root.left.right = newNode(14);
    root.right = newNode(18);
    root.right.left = newNode(17);
    root.right.right = newNode(21);
  
    Console.WriteLine("{0} "
          countOddXor(root, null));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

5


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Improved By : princiraj1992, 29AjayKumar