# 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**:

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

Below is the implementation of the above approach:

`// 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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**Output:**

5

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