# Print cousins of a given node in Binary Tree

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea to first find level of given node using the approach discussed here. Once we have found level, we can print all nodes at a given level using the approach discussed here. 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 the implementation of above idea.

## C++

 `// C++ program to print cousins of a node ` `#include ` `using` `namespace` `std; ` ` `  `// 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) ` `            ``cout << root->left->data << ``" "``; ` `        ``if` `(root->right) ` `            ``cout << 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 Code ` `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; ` `} ` ` `  `// This code is contributed ` `// by Akanksha Rai `

## C

 `// C program to print cousins of a node ` `#include ` `#include ` ` `  `// 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; ` `} `

## Java

 `// Java program to print cousins of a node  ` `class` `GfG {  ` ` `  `// A Binary Tree Node  ` `static` `class` `Node  ` `{  ` `    ``int` `data;  ` `    ``Node left, right;  ` `} ` ` `  `// A utility function to create a new Binary  ` `// Tree Node  ` `static` `Node newNode(``int` `item)  ` `{  ` `    ``Node temp = ``new` `Node();  ` `    ``temp.data = item;  ` `    ``temp.left = ``null``; ` `    ``temp.right = ``null``;  ` `    ``return` `temp;  ` `}  ` ` `  `/* It returns level of the node if it is present  ` `in tree, otherwise returns 0.*/` `static` `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 */` `static` `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 != ``null``)  ` `        ``System.out.print(root.left.data + ``" "``);  ` `        ``if` `(root.right != ``null``)  ` `        ``System.out.print(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  ` `static` `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  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``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);  ` `}  ` `}  `

## Python3

 `# Python3 program to print cousins of a node  ` ` `  `# A utility function to create a new  ` `# Binary Tree Node  ` `class` `newNode: ` `    ``def` `__init__(``self``, item): ` `        ``self``.data ``=` `item  ` `        ``self``.left ``=` `self``.right ``=` `None` ` `  `# It returns level of the node if it is  ` `# present in tree, otherwise returns 0. ` `def` `getLevel(root, node, level): ` `     `  `    ``# base cases  ` `    ``if` `(root ``=``=` `None``): ` `        ``return` `0` `    ``if` `(root ``=``=` `node): ` `        ``return` `level  ` ` `  `    ``# If node is present in left subtree  ` `    ``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``) ` ` `  `# Prnodes at a given level such that  ` `# sibling of node is not printed if  ` `# it exists  ` `def` `printGivenLevel(root, node, level): ` `     `  `    ``# Base cases  ` `    ``if` `(root ``=``=` `None` `or` `level < ``2``):  ` `        ``return` ` `  `    ``# If current node is parent of a  ` `    ``# node with given level  ` `    ``if` `(level ``=``=` `2``): ` `        ``if` `(root.left ``=``=` `node ``or` `            ``root.right ``=``=` `node):  ` `            ``return` `        ``if` `(root.left):  ` `            ``print``(root.left.data, end ``=` `" "``)  ` `        ``if` `(root.right): ` `            ``print``(root.right.data, end ``=` `" "``) ` ` `  `    ``# Recur for left and right subtrees  ` `    ``elif` `(level > ``2``): ` `        ``printGivenLevel(root.left, node, level ``-` `1``)  ` `        ``printGivenLevel(root.right, node, level ``-` `1``) ` ` `  `# This function prints cousins of a given node  ` `def` `printCousins(root, node): ` `     `  `    ``# Get level of given node  ` `    ``level ``=` `getLevel(root, node, ``1``)  ` ` `  `    ``# Prnodes of given level.  ` `    ``printGivenLevel(root, node, level) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``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) ` `     `  `# This code is contributed by PranchalK `

## C#

 `// C# program to print cousins of a node  ` `using` `System; ` ` `  `public` `class` `GfG  ` `{  ` ` `  `// A Binary Tree Node  ` `class` `Node  ` `{  ` `    ``public` `int` `data;  ` `    ``public` `Node left, right;  ` `} ` ` `  `// A utility function to create   ` `// a new Binary Tree Node  ` `static` `Node newNode(``int` `item)  ` `{  ` `    ``Node temp = ``new` `Node();  ` `    ``temp.data = item;  ` `    ``temp.left = ``null``; ` `    ``temp.right = ``null``;  ` `    ``return` `temp;  ` `}  ` ` `  `/* It returns level of the node ` `if it is present in tree, ` ` ``otherwise returns 0.*/` `static` `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 */` `static` `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 != ``null``)  ` `            ``Console.Write(root.left.data + ``" "``);  ` `        ``if` `(root.right != ``null``)  ` `            ``Console.Write(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  ` `static` `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 code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``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);  ` `}  ` `} ` ` `  `// This code is contributed Rajput-Ji  `

Output :

`6 7`

Time Complexity : O(n)

Can we solve this problem using single traversal? Please refer below article
Print cousins of a given node in Binary Tree | Single Traversal

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