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

Given a binary tree and the two nodes say ‘a’ and ‘b’, determine whether two given 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.

A solution in Set-1 that finds whether given nodes are cousins or not by performing three traversals of binary tree has been discussed. The problem can be solved by performing level order traversal. The idea is to use a queue to perform level order traversal, in which each queue element is a pair of node and parent of that node. For each node visited in level order traversal, check if that node is either first given node or second given node. If any node is found store parent of that node. While performing level order traversal, one level is traversed at a time. If both nodes are found in given level, then their parent values are compared to check if they are siblings or not. If one node is found in given level and another is not found, then given nodes are not cousins.

Steps to solve this problem:

1. Check if root is null than return false.

2. Declare two pointers parA=null and parB=null.

3. Declare a queue pair q of pointer to a node both.

4. Declare a pointer to a new node tmp with value -1.

5. Declare a pair ele of pointer to a node both.

6. Push (root,tmp) in q and declare a variable levsize.

7. While q is not empty:

    *Update levsize =q.size

    *While levsize is not zero:

        *Update ele=q.front and pop the element after that.

        *Check if ele.first->data is equal to a->data than parA=ele.second

        *Check if ele.first->data is equal to b->data than parB=ele.second

        *Check if ele.first->left is not null than push (ele.first->left,ele.first) in q.

        *Check if ele.first->right is not null than push (ele.first->right,ele.first) in q.

        *Decrement levsize.

        *Check parA logical AND parB is true than break.

    *Check if parA logical AND parB is true than return parA not equal to parB.

    *Check if (parA && !parB)||(parB && !parA) is true than return false.

8. Return false.

Below is the implementation of above approach:

C++

// CPP program to check if two Nodes in
// a binary tree are cousins
// using level-order traversals
#include <bits/stdc++.h>
using namespace std;

// 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;
}

// Returns true if a and b are cousins,
// otherwise false.
bool isCousin(Node* root, Node* a, Node* b)
{
    if (root == NULL)
        return false;

    // To store parent of node a.
    Node* parA = NULL;

    // To store parent of node b.
    Node* parB = NULL;

    // queue to perform level order
    // traversal. Each element of
    // queue is a pair of node and
    // its parent.
    queue<pair<Node*, Node*> > q;

    // Dummy node to act like parent
    // of root node.
    Node* tmp = newNode(-1);

    // To store front element of queue.
    pair<Node*, Node*> ele;

    // Push root to queue.
    q.push(make_pair(root, tmp));
    int levSize;

    while (!q.empty()) {

        // find number of elements in
        // current level.
        levSize = q.size();
        while (levSize) {

            ele = q.front();
            q.pop();

            // check if current node is node a
            // or node b or not.
            if (ele.first->data == a->data) {
                parA = ele.second;
            }

            if (ele.first->data == b->data) {
                parB = ele.second;
            }

            // push children of current node
            // to queue.
            if (ele.first->left) {
                q.push(make_pair(ele.first->left, ele.first));
            }

            if (ele.first->right) {
                q.push(make_pair(ele.first->right, ele.first));
            }

            levSize--;

            // If both nodes are found in
            // current level then no need
            // to traverse current level further.
            if (parA && parB)
                break;
        }

        // Check if both nodes are siblings
        // or not.
        if (parA && parB) {
            return parA != parB;
        }

        // If one node is found in current level
        // and another is not found, then
        // both nodes are not cousins.
        if ((parA && !parB) || (parB && !parA)) {
            return false;
        }
    }

    return false;
}
// Driver Code
int main()
{
    /*
            1
           /  \
          2    3
         / \  / \
        4   5 6  7
             \ \
             15 8
    */

    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 Nodes in
// a binary tree are cousins
// using level-order traversals
import java.util.*;
import javafx.util.Pair;

// User defined node class
class Node
{
    int data;
    Node left, right;

    // Constructor to create a new tree node
    Node(int item)
    {
        data = item;
        left = right = null;
    }
}

class BinaryTree
{
    Node root;

    // Returns true if a and b are cousins,
    // otherwise false.
    boolean isCousin(Node node, Node a, Node b)
    {
        if(node == null)
            return false;

        // To store parent of node a.
        Node parA = null;

        // To store parent of node b.
        Node parB = null;

        // queue to perform level order
        // traversal. Each element of
        // queue is a pair of node and
        // its parent.
        Queue<Pair <Node,Node>> q = new LinkedList<> ();

        // Dummy node to act like parent
        // of root node.
        Node tmp = new Node(-1);

        // To store front element of queue.
        Pair<Node, Node> ele;

        // Push root to queue.
        q.add(new Pair <Node,Node> (node, tmp));

        int levelSize;

        while(!q.isEmpty())
        {

            // find number of elements in
            // current level.
            levelSize = q.size();
            while(levelSize != 0)
            {
                ele = q.peek();
                q.remove();

                // check if current node is node a
                // or node b or not.
                if(ele.getKey().data == a.data)
                    parA = ele.getValue();

                if(ele.getKey().data == b.data)
                    parB = ele.getValue();

                // push children of current node
                // to queue.
                if(ele.getKey().left != null)
                    q.add(new Pair<Node, Node>(ele.getKey().left, ele.getKey()));

                if(ele.getKey().right != null)
                    q.add(new Pair<Node, Node>(ele.getKey().right, ele.getKey()));

                levelSize--;

                // If both nodes are found in
                // current level then no need
                // to traverse current level further.
                if(parA != null && parB != null)
                    break;
            }

            // Check if both nodes are siblings
            // or not.
            if(parA != null && parB != null)
                return parA != parB;

            // If one node is found in current level
            // and another is not found, then
            // both nodes are not cousins.
            if ((parA!=null && parB==null) || (parB!=null && parA==null))
                return false;
        }

        return false;
    }

    // Driver code
    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.right.right;
        Node2 = tree.root.right.left.right;
        if (tree.isCousin(tree.root, Node1, Node2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by shubham96301

Python3

# Python3 program to check if two
# Nodes in a binary tree are cousins
# using level-order traversals

# A Binary Tree Node
class Node:

    def __init__(self, item):
        self.data = item
        self.left = None
        self.right = None

# Returns True if a and b
# are cousins, otherwise False.
def isCousin(root, a, b):

    if root == None:
        return False

    # To store parent of node a.
    parA = None

    # To store parent of node b.
    parB = None

    # queue to perform level order
    # traversal. Each element of queue
    # is a pair of node and its parent.
    q = []

    # Dummy node to act like
    # parent of root node.
    tmp = Node(-1)

    # Push root to queue.
    q.append((root, tmp))

    while len(q) > 0:

        # find number of elements in
        # current level.
        levSize = len(q)
        while levSize:

            ele = q.pop(0)

            # check if current node is
            # node a or node b or not.
            if ele[0].data == a.data:
                parA = ele[1]

            if ele[0].data == b.data:
                parB = ele[1]

            # push children of
            # current node to queue.
            if ele[0].left:
                q.append((ele[0].left, ele[0]))

            if ele[0].right:
                q.append((ele[0].right, ele[0]))
            levSize -= 1

            # If both nodes are found in
            # current level then no need
            # to traverse current level further.
            if parA and parB:
                break

        # Check if both nodes
        # are siblings or not.
        if parA and parB:
            return parA != parB

        # If one node is found in current level
        # and another is not found, then
        # both nodes are not cousins.
        if (parA and not parB) or (parB and not parA):
            return False

    return False

# Driver Code
if __name__ == '__main__':

    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.left
    Node2 = root.right.right

    if isCousin(root, Node1, Node2):
        print('Yes')
    else:
        print('No')

# This code is contributed by Rituraj Jain

C#

// C# program to check if two Nodes in
// a binary tree are cousins
// using level-order traversals
using System;
using System.Collections.Generic;

// User defined node class
public class Node
{
    public int data;
    public Node left, right;

    // Constructor to create a new tree node
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}
// User defined pair class
public class Pair
{

    public Node first, second;

    // Constructor to create a new tree node
    public Pair(Node first, Node second)
    {
        this.first = first;
        this.second = second;
    }
}

class BinaryTree
{
    Node root;

    // Returns true if a and b are cousins,
    // otherwise false.
    Boolean isCousin(Node node, Node a, Node b)
    {
        if(node == null)
            return false;

        // To store parent of node a.
        Node parA = null;

        // To store parent of node b.
        Node parB = null;

        // queue to perform level order
        // traversal. Each element of
        // queue is a pair of node and
        // its parent.
        Queue<Pair > q = new Queue<Pair > ();

        // Dummy node to act like parent
        // of root node.
        Node tmp = new Node(-1);

        // To store front element of queue.
        Pair ele;

        // Push root to queue.
        q.Enqueue(new Pair (node, tmp));

        int levelSize;

        while(q.Count>0)
        {

            // find number of elements in
            // current level.
            levelSize = q.Count;
            while(levelSize != 0)
            {
                ele = q.Peek();
                q.Dequeue();

                // check if current node is node a
                // or node b or not.
                if(ele.first.data == a.data)
                    parA = ele.second;

                if(ele.first.data == b.data)
                    parB = ele.second;

                // push children of current node
                // to queue.
                if(ele.first.left != null)
                    q.Enqueue(new Pair(ele.first.left, ele.first));

                if(ele.first.right != null)
                    q.Enqueue(new Pair(ele.first.right, ele.first));

                levelSize--;

                // If both nodes are found in
                // current level then no need
                // to traverse current level further.
                if(parA != null && parB != null)
                    break;
            }

            // Check if both nodes are siblings
            // or not.
            if(parA != null && parB != null)
                return parA != parB;

            // If one node is found in current level
            // and another is not found, then
            // both nodes are not cousins.
            if ((parA != null && parB == null) || (parB != null && parA == null))
                return false;
        }

        return false;
    }

    // Driver code
    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.right.right;
        Node2 = tree.root.right.left.right;
        if (tree.isCousin(tree.root, Node1, Node2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by Arnab Kundu

Javascript

<script>

// JavaScript program to check if two Nodes in
// a binary tree are cousins
// using level-order traversals

// User defined node class
class Node
{

    // Constructor to create a new tree node
    constructor(item)
    {
        this.data = item;
        this.left = null;
        this.right = null;
    }

}
// User defined pair class
class Pair
{

    // Constructor to create a new tree node
    constructor(first, second)
    {
        this.first = first;
        this.second = second;
    }

}

var root = null;
// Returns true if a and b are cousins,
// otherwise false.
function isCousin(node, a, b)
{
    if(node == null)
        return false;

    // To store parent of node a.
    var parA = null;
    // To store parent of node b.
    var parB = null;
    // queue to perform level order
    // traversal. Each element of
    // queue is a pair of node and
    // its parent.
    var q = [];
    // Dummy node to act like parent
    // of root node.
    var tmp = new Node(-1);
    // To store front element of queue.
    var ele;
    // Push root to queue.
    q.push(new Pair (node, tmp));
    var levelSize;
    while(q.length>0)
    {
        // find number of elements in
        // current level.
        levelSize = q.length;
        while(levelSize != 0)
        {
            ele = q[0];
            q.shift();
            // check if current node is node a
            // or node b or not.
            if(ele.first.data == a.data)
                parA = ele.second;
            if(ele.first.data == b.data)
                parB = ele.second;
            // push children of current node
            // to queue.
            if(ele.first.left != null)
                q.push(new Pair(ele.first.left, ele.first));
            if(ele.first.right != null)
                q.push(new Pair(ele.first.right, ele.first));
            levelSize--;
            // If both nodes are found in
            // current level then no need
            // to traverse current level further.
            if(parA != null && parB != null)
                break;
        }
        // Check if both nodes are siblings
        // or not.
        if(parA != null && parB != null)
            return parA != parB;
        // If one node is found in current level
        // and another is not found, then
        // both nodes are not cousins.
        if ((parA != null && parB == null) || (parB != null
        && parA == null))
            return false;
    }
    return false;
}
// Driver code
root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.left.right.right = new Node(15);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.right.left.right = new Node(8);
var Node1, Node2;
Node1 = root.left.right.right;
Node2 = root.right.left.right;
if (isCousin(root, Node1, Node2))
    document.write("Yes");
else
    document.write("No");


</script>

Output

Yes

Complexity Analysis:

• Time Complexity: O(n) 
• Auxiliary Space: O(n)