# Check if two nodes are cousins in a Binary Tree

Given the binary Tree and the two nodes say ‘a’ and ‘b’, determine whether the two nodes are cousins of each other or not.

Two nodes are cousins of each other if they are at same level and have different parents.

Example:

```     6
/   \
3     5
/ \   / \
7   8 1   3
Say two node be 7 and 1, result is TRUE.
Say two nodes are 3 and 5, result is FALSE.
Say two nodes are 7 and 5, result is FALSE.```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to find level of one of the nodes. Using the found level, check if ‘a’ and ‘b’ are at this level. If ‘a’ and ‘b’ are at given level, then finally check if they are not children of same parent.

Following is the implementation of the above approach.

## C

 `// C program to check if two Nodes in a binary tree are cousins ` `#include ` `#include ` ` `  `// A Binary Tree Node ` `struct` `Node ` `{ ` `    ``int` `data; ` `    ``struct` `Node *left, *right; ` `}; ` ` `  `// A utility function to create a new Binary Tree Node ` `struct` `Node *newNode(``int` `item) ` `{ ` `    ``struct` `Node *temp =  (``struct` `Node *)``malloc``(``sizeof``(``struct` `Node)); ` `    ``temp->data = item; ` `    ``temp->left = temp->right = NULL; ` `    ``return` `temp; ` `} ` ` `  `// Recursive function to check if two Nodes are siblings ` `int` `isSibling(``struct` `Node *root, ``struct` `Node *a, ``struct` `Node *b) ` `{ ` `    ``// Base case ` `    ``if` `(root==NULL)  ``return` `0; ` ` `  `    ``return` `((root->left==a && root->right==b)|| ` `            ``(root->left==b && root->right==a)|| ` `            ``isSibling(root->left, a, b)|| ` `            ``isSibling(root->right, a, b)); ` `} ` ` `  `// Recursive function to find level of Node 'ptr' in a binary tree ` `int` `level(``struct` `Node *root, ``struct` `Node *ptr, ``int` `lev) ` `{ ` `    ``// base cases ` `    ``if` `(root == NULL) ``return` `0; ` `    ``if` `(root == ptr)  ``return` `lev; ` ` `  `    ``// Return level if Node is present in left subtree ` `    ``int` `l = level(root->left, ptr, lev+1); ` `    ``if` `(l != 0)  ``return` `l; ` ` `  `    ``// Else search in right subtree ` `    ``return` `level(root->right, ptr, lev+1); ` `} ` ` `  ` `  `// Returns 1 if a and b are cousins, otherwise 0 ` `int` `isCousin(``struct` `Node *root, ``struct` `Node *a, ``struct` `Node *b) ` `{ ` `    ``//1. The two Nodes should be on the same level in the binary tree. ` `    ``//2. The two Nodes should not be siblings (means that they should ` `    ``// not have the same parent Node). ` `    ``if` `((level(root,a,1) == level(root,b,1)) && !(isSibling(root,a,b))) ` `        ``return` `1; ` `    ``else` `return` `0; ` `} ` ` `  `// Driver Program to test above functions ` `int` `main() ` `{ ` `    ``struct` `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); ` ` `  `    ``struct` `Node *Node1,*Node2; ` `    ``Node1 = root->left->left; ` `    ``Node2 = root->right->right; ` ` `  `    ``isCousin(root,Node1,Node2)? ``puts``(``"Yes"``): ``puts``(``"No"``); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to check if two binary tree are cousins ` `class` `Node ` `{ ` `    ``int` `data; ` `    ``Node left, right; ` ` `  `    ``Node(``int` `item) ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `class` `BinaryTree ` `{ ` `    ``Node root; ` ` `  `    ``// Recursive function to check if two Nodes are ` `    ``// siblings ` `    ``boolean` `isSibling(Node node, Node a, Node b) ` `    ``{ ` `        ``// Base case ` `        ``if` `(node == ``null``) ` `            ``return` `false``; ` ` `  `        ``return` `((node.left == a && node.right == b) || ` `                ``(node.left == b && node.right == a) || ` `                ``isSibling(node.left, a, b) || ` `                ``isSibling(node.right, a, b)); ` `    ``} ` ` `  `    ``// Recursive function to find level of Node 'ptr' in ` `    ``// a binary tree ` `    ``int` `level(Node node, Node ptr, ``int` `lev) ` `    ``{ ` `        ``// base cases ` `        ``if` `(node == ``null``) ` `            ``return` `0``; ` ` `  `        ``if` `(node == ptr) ` `            ``return` `lev; ` ` `  `        ``// Return level if Node is present in left subtree ` `        ``int` `l = level(node.left, ptr, lev + ``1``); ` `        ``if` `(l != ``0``) ` `            ``return` `l; ` ` `  `        ``// Else search in right subtree ` `        ``return` `level(node.right, ptr, lev + ``1``); ` `    ``} ` ` `  `    ``// Returns 1 if a and b are cousins, otherwise 0 ` `    ``boolean` `isCousin(Node node, Node a, Node b) ` `    ``{ ` `        ``// 1. The two Nodes should be on the same level ` `        ``//       in the binary tree. ` `        ``// 2. The two Nodes should not be siblings (means ` `        ``//    that they should not have the same parent ` `        ``//    Node). ` `        ``return` `((level(node, a, ``1``) == level(node, b, ``1``)) && ` `                ``(!isSibling(node, a, b))); ` `    ``} ` ` `  `    ``//Driver program to test above functions ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``tree.root = ``new` `Node(``1``); ` `        ``tree.root.left = ``new` `Node(``2``); ` `        ``tree.root.right = ``new` `Node(``3``); ` `        ``tree.root.left.left = ``new` `Node(``4``); ` `        ``tree.root.left.right = ``new` `Node(``5``); ` `        ``tree.root.left.right.right = ``new` `Node(``15``); ` `        ``tree.root.right.left = ``new` `Node(``6``); ` `        ``tree.root.right.right = ``new` `Node(``7``); ` `        ``tree.root.right.left.right = ``new` `Node(``8``); ` ` `  `        ``Node Node1, Node2; ` `        ``Node1 = tree.root.left.left; ` `        ``Node2 = tree.root.right.right; ` `        ``if` `(tree.isCousin(tree.root, Node1, Node2)) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `            ``System.out.println(``"No"``); ` `    ``} ` `} ` ` `  `// This code has been contributed by Mayank Jaiswal `

## Python

 `# Python program to check if two nodes in a binary  ` `# tree are cousins ` ` `  `# A Binary Tree Node ` `class` `Node: ` `     `  `    ``# Constructor to create a new Binary Tree ` `    ``def` `__init__(``self``, data): ` `        ``self``.data ``=` `data ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  `def` `isSibling(root, a , b): ` ` `  `    ``# Base Case ` `    ``if` `root ``is` `None``: ` `        ``return` `0` ` `  `    ``return` `((root.left ``=``=` `a ``and` `root.right ``=``=``b) ``or`  `            ``(root.left ``=``=` `b ``and` `root.right ``=``=` `a)``or` `            ``isSibling(root.left, a, b) ``or` `            ``isSibling(root.right, a, b)) ` ` `  `# Recursive function to find level of Node 'ptr' in  ` `# a binary tree ` `def` `level(root, ptr, lev): ` `     `  `    ``# Base Case  ` `    ``if` `root ``is` `None` `: ` `        ``return` `0`  `    ``if` `root ``=``=` `ptr:  ` `        ``return` `lev ` ` `  `    ``# Return level if Node is present in left subtree ` `    ``l ``=` `level(root.left, ptr, lev``+``1``) ` `    ``if` `l !``=` `0``: ` `        ``return` `l ` ` `  `    ``# Else search in right subtree ` `    ``return` `level(root.right, ptr, lev``+``1``) ` ` `  ` `  `# Returns 1 if a and b are cousins, otherwise 0 ` `def` `isCousin(root,a, b): ` `     `  `    ``# 1. The two nodes should be on the same level in  ` `    ``# the binary tree ` `    ``# The two nodes should not be siblings(means that  ` `    ``# they should not have the smae parent node ` ` `  `    ``if` `((level(root,a,``1``) ``=``=` `level(root, b, ``1``)) ``and`  `            ``not` `(isSibling(root, a, b))): ` `        ``return` `1` `    ``else``: ` `        ``return` `0`  ` `  ` `  `# Driver program to test above function ` `root ``=` `Node(``1``) ` `root.left ``=` `Node(``2``) ` `root.right ``=` `Node(``3``) ` `root.left.left ``=` `Node(``4``) ` `root.left.right ``=` `Node(``5``) ` `root.left.right.right ``=` `Node(``15``) ` `root.right.left ``=` `Node(``6``) ` `root.right.right ``=` `Node(``7``) ` `root.right.left.right ``=` `Node(``8``) ` ` `  `node1 ``=` `root.left.right ` `node2 ``=` `root.right.right  ` ` `  `print` `"Yes"` `if` `isCousin(root, node1, node2) ``=``=` `1` `else` `"No"` ` `  `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) `

## C#

 `// C# program to check if two binary  ` `// tree are cousins  ` `using` `System; ` ` `  `public` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node left, right; ` ` `  `    ``public` `Node(``int` `item) ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `class` `GFG ` `{ ` `public` `Node root; ` ` `  `// Recursive function to check if  ` `// two Nodes are siblings  ` `public` `virtual` `bool` `isSibling(Node node,  ` `                              ``Node a, Node b) ` `{ ` `    ``// Base case  ` `    ``if` `(node == ``null``) ` `    ``{ ` `        ``return` `false``; ` `    ``} ` ` `  `    ``return` `((node.left == a && node.right == b) ||  ` `            ``(node.left == b && node.right == a) ||  ` `                     ``isSibling(node.left, a, b) ||  ` `                     ``isSibling(node.right, a, b)); ` `} ` ` `  `// Recursive function to find level  ` `// of Node 'ptr' in a binary tree  ` `public` `virtual` `int` `level(Node node, Node ptr, ``int` `lev) ` `{ ` `    ``// base cases  ` `    ``if` `(node == ``null``) ` `    ``{ ` `        ``return` `0; ` `    ``} ` ` `  `    ``if` `(node == ptr) ` `    ``{ ` `        ``return` `lev; ` `    ``} ` ` `  `    ``// Return level if Node is present  ` `    ``// in left subtree  ` `    ``int` `l = level(node.left, ptr, lev + 1); ` `    ``if` `(l != 0) ` `    ``{ ` `        ``return` `l; ` `    ``} ` ` `  `    ``// Else search in right subtree  ` `    ``return` `level(node.right, ptr, lev + 1); ` `} ` ` `  `// Returns 1 if a and b are cousins,  ` `// otherwise 0  ` `public` `virtual` `bool` `isCousin(Node node,  ` `                             ``Node a, Node b) ` `{ ` `    ``// 1. The two Nodes should be on the  ` `    ``//      same level in the binary tree.  ` `    ``// 2. The two Nodes should not be siblings  ` `    ``// (means that they should not have the  ` `    ``//  same parent Node).  ` `    ``return` `((level(node, a, 1) == level(node, b, 1)) &&  ` `                            ``(!isSibling(node, a, b))); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``GFG tree = ``new` `GFG(); ` `    ``tree.root = ``new` `Node(1); ` `    ``tree.root.left = ``new` `Node(2); ` `    ``tree.root.right = ``new` `Node(3); ` `    ``tree.root.left.left = ``new` `Node(4); ` `    ``tree.root.left.right = ``new` `Node(5); ` `    ``tree.root.left.right.right = ``new` `Node(15); ` `    ``tree.root.right.left = ``new` `Node(6); ` `    ``tree.root.right.right = ``new` `Node(7); ` `    ``tree.root.right.left.right = ``new` `Node(8); ` ` `  `    ``Node Node1, Node2; ` `    ``Node1 = tree.root.left.left; ` `    ``Node2 = tree.root.right.right; ` `    ``if` `(tree.isCousin(tree.root, Node1, Node2)) ` `    ``{ ` `        ``Console.WriteLine(``"Yes"``); ` `    ``} ` `    ``else` `    ``{ ` `        ``Console.WriteLine(``"No"``); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by Shrikant13 `

Output:

` Yes `

Time Complexity of the above solution is O(n) as it does at most three traversals of binary tree.

Check if two nodes are cousins in a Binary Tree | Set-2

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Improved By : shrikanth13, Akanksha_Rai

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