# Print cousins of a given node in Binary Tree

Datetime:2016-08-23 01:19:41         Topic:         Share        Original >>

Given a binary tree and a node, print all cousins of given node. Note that siblings should not be printed.

Example:

```Input : root of below tree
1
/   \
2     3
/   \  /  \
4    5  6   7
and pointer to a node say 5.

Output : 6, 7```

We strongly recommend you to minimize your browser and try this yourself first.

The idea to first find level of given node using the approach discussedhere. Once we have found level, we can print all nodes at a given level using the approach discussedhere. The only thing to take care of is, sibling should not be printed. To handle this, we change the printing function to first check for sibling and print node only if it is not sibling.

Below is C++ implementation of above idea.

```// C program to print cousins of a node
#include <stdio.h>
#include <stdlib.h>

// A Binary Tree Node
struct Node
{
int data;
Node *left, *right;
};

// A utility function to create a new Binary
// Tree Node
Node *newNode(int item)
{
Node *temp =  new Node;
temp->data = item;
temp->left = temp->right = NULL;
return temp;
}

/* It returns level of the node if it is present
in tree, otherwise returns 0.*/
int getLevel(Node *root, Node *node, int level)
{
// base cases
if (root == NULL)
return 0;
if (root == node)
return level;

// If node is present in left subtree
int downlevel = getLevel(root->left, node, level+1);
if (downlevel != 0)
return downlevel;

// If node is not present in left subtree
return getLevel(root->right, node, level+1);
}

/* Print nodes at a given level such that sibling of
node is not printed if it exists  */
void printGivenLevel(Node* root, Node *node, int level)
{
// Base cases
if (root == NULL || level < 2)
return;

// If current node is parent of a node with
// given level
if (level == 2)
{
if (root->left == node || root->right == node)
return;
if (root->left)
printf("%d ", root->left->data);
if (root->right)
printf("%d ", root->right->data);
}

// Recur for left and right subtrees
else if (level > 2)
{
printGivenLevel(root->left, node, level-1);
printGivenLevel(root->right, node, level-1);
}
}

// This function prints cousins of a given node
void printCousins(Node *root, Node *node)
{
// Get level of given node
int level = getLevel(root, node, 1);

// Print nodes of given level.
printGivenLevel(root, node, level);
}

// Driver Program to test above functions
int main()
{
Node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->left->right->right = newNode(15);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->right->left->right = newNode(8);

printCousins(root, root->left->right);

return 0;
}```

Output :

Time Complexity : O(n)