Find the cousins of a given element in an N-ary tree

Last Updated : 31 Jan, 2022

Given an N-array tree root and an integer K, the task is to print all the cousins of node K.

Note: Two nodes are considered to be cousins if they have the same depth (are on the same level) and have different parents.

Examples:

Consider the below tree:

Input: K = 39
Output: 88 98 61 74 17 72 19

Input: K = 17
Output: 88 98 61 74 39

Input: K = 90
Output: NA

Approach: The idea is to do a level order traversal. During the traversal, if we find a node whose child is equal to the given element, then we will not push the children of this node. We will push the children of the other nodes and the inner loop will end when all the elements of that level are traversed. Follow the steps below to solve the problem:

If root equals null, then return.
Initialize the queue q[] and push root into the queue q[].
Initialize the boolean variable found as false.
Initialize the variables qsize as 0 and node temp.
Iterate over the while loop till q[] is not empty and node is not found and perform the following tasks:
- Set size qsize as the size of the queue q[].
- Iterate over the while loop till qsize is greater than 0 and perform the following tasks:
  - Set tempp as the front of the queue q[].
  - De queue from the queue q[].
  - If found equals true, then push all it’s children into the queue q[].
  - Iterate over the range [0, temp->child.size()) using the variable i and perform the following tasks:
    - If the child is not null and it’s key equals value, then set the value of found as true.
  - If found is false, then push all it’s children into the queue q[].
- Decrease the value of qsize by 1.
If found is false, then print “Not Possible.”
Else, Initialize the variables qsize as the size of the queue q[].
If qsize equals 0 then print “No Cousins”.
Else, print all the elements of the queue q[].

Below is the implementation of the above approach

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a node of N-ary tree
struct Node {
    int key;
    vector<Node*> child;
};
 
// New node creation
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    return temp;
}
 
// Function to find the cousins of a
// given node in an N-array tree
void printCousins(Node* root, int value)
{
 
    // Base case
    if (root == NULL)
        return;
 
    queue<Node*> q;
    q.push(root);
 
    // If we find the node
    // with value as the key
    bool found = false;
 
    int qsize = 0;
    Node* tempp;
 
    while (!q.empty() && !found) {
        qsize = q.size();
 
        while (qsize) {
 
            // Storing the current node
            tempp = q.front();
            q.pop();
 
            // If we have already found
            // the value as child of a node,
            // we need to insert children of other
            // node of same level in the queue
            if (found == true) {
                for (int i = 0; i < tempp->child.size();
                     i++) {
                    if (tempp->child[i] != NULL)
                        q.push(tempp->child[i]);
                }
            }
 
            // If value is child of tempp node
            for (int i = 0; i < tempp->child.size(); i++)
                if (tempp->child[i] != NULL
                    && tempp->child[i]->key == value)
                    found = true;
 
            // If value is not the child of tempp node
            // then insert all the children
            // of the tempp node
            if (found == false) {
                for (int i = 0; i < tempp->child.size();
                     i++) {
                    if (tempp->child[i] != NULL)
                        q.push(tempp->child[i]);
                }
            }
 
            qsize--;
        }
    }
 
    if (found) {
 
        // Queue will contain the cousins
        qsize = q.size();
 
        if (qsize == 0)
            cout << "NA";
        for (int i = 0; i < qsize; i++) {
            tempp = q.front();
            q.pop();
            cout << tempp->key << " ";
        }
    }
    else {
 
        // When value  is not in the tree
        cout << "Not Possible";
    }
    cout << "\n";
    return;
}
 
// Driver Code
int main()
{
 
    Node* root = newNode(10);
    (root->child).push_back(newNode(77));
    (root->child).push_back(newNode(90));
    (root->child).push_back(newNode(35));
    (root->child).push_back(newNode(19));
    (root->child[0]->child).push_back(newNode(88));
    (root->child[0]->child).push_back(newNode(98));
    (root->child[0]->child[1]->child)
        .push_back(newNode(76));
    (root->child[0]->child[1]->child)
        .push_back(newNode(20));
    (root->child[1]->child).push_back(newNode(61));
    (root->child[1]->child).push_back(newNode(74));
    (root->child[2]->child).push_back(newNode(39));
    (root->child[3]->child).push_back(newNode(17));
    (root->child[3]->child).push_back(newNode(72));
    (root->child[3]->child).push_back(newNode(19));
 
    // Find the cousins of value
    int value = 39;
    printCousins(root, value);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
  // Structure of a node of N-ary tree
  static class Node {
    int key;
    Vector<Node> child = new Vector<Node>();
  };
 
  // New node creation
  static Node newNode(int key)
  {
    Node temp = new Node();
    temp.key = key;
    return temp;
  }
 
  // Function to find the cousins of a
  // given node in an N-array tree
  static void printCousins(Node root, int value)
  {
 
    // Base case
    if (root == null)
      return;
 
    Queue<Node> q = new LinkedList<GFG.Node>();
    q.add(root);
 
    // If we find the node
    // with value as the key
    boolean found = false;
 
    int qsize = 0;
    Node tempp;
 
    while (!q.isEmpty() && !found) {
      qsize = q.size();
 
      while (qsize > 0) {
 
        // Storing the current node
        tempp = q.peek();
        q.remove();
 
        // If we have already found
        // the value as child of a node,
        // we need to insert children of other
        // node of same level in the queue
        if (found == true) {
          for (int i = 0; i < tempp.child.size();
               i++) {
            if (tempp.child.get(i) != null)
              q.add(tempp.child.get(i));
          }
        }
 
        // If value is child of tempp node
        for (int i = 0; i < tempp.child.size(); i++)
          if (tempp.child.get(i) != null
              && tempp.child.get(i).key == value)
            found = true;
 
        // If value is not the child of tempp node
        // then insert all the children
        // of the tempp node
        if (found == false) {
          for (int i = 0; i < tempp.child.size();
               i++) {
            if (tempp.child.get(i) != null)
              q.add(tempp.child.get(i));
          }
        }
 
        qsize--;
      }
    }
 
    if (found) {
 
      // Queue will contain the cousins
      qsize = q.size();
 
      if (qsize == 0)
        System.out.print("NA");
      for (int i = 0; i < qsize; i++) {
        tempp = q.peek();
        q.remove();
        System.out.print(tempp.key+ " ");
      }
    }
    else {
 
      // When value  is not in the tree
      System.out.print("Not Possible");
    }
    System.out.print("\n");
    return;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    Node root = newNode(10);
    (root.child).add(newNode(77));
    (root.child).add(newNode(90));
    (root.child).add(newNode(35));
    (root.child).add(newNode(19));
    (root.child.get(0).child).add(newNode(88));
    (root.child.get(0).child).add(newNode(98));
    (root.child.get(0).child.get(1).child)
    .add(newNode(76));
    (root.child.get(0).child.get(1).child)
    .add(newNode(20));
    (root.child.get(1).child).add(newNode(61));
    (root.child.get(1).child).add(newNode(74));
    (root.child.get(2).child).add(newNode(39));
    (root.child.get(3).child).add(newNode(17));
    (root.child.get(3).child).add(newNode(72));
    (root.child.get(3).child).add(newNode(19));
 
    // Find the cousins of value
    int value = 39;
    printCousins(root, value);
 
  }
}
 
// This code is contributed by shikhasingrajput

Python3

# Python code for the above approach 
 
# Structure of a node of N-ary tree
class Node:
    def __init__ (self, k):
        self.key = k;
        self.child = [];
 
# node creation
 
# Function to find the cousins of a
# given node in an N-array tree
def printCousins(root, value):
 
    # Base case
    if (root == None):
        return;
 
    q = [];
    q.append(root);
 
    # If we find the node
    # with value as the key
    found = False;
 
    qsize = 0;
    tempp = None
 
    while (len(q) != 0 and found != 1):
        qsize = len(q);
 
        while (qsize != 0):
 
            # Storing the current node
            tempp = q[0];
            q.pop(0);
 
            # If we have already found
            # the value as child of a node,
            # we need to insert children of other
            # node of same level in the queue
            if (found == True):
                for i in range(len(tempp.child)):
                    if (tempp.child[i] != None):
                        q.append(tempp.child[i]);
 
            # If value is child of tempp node
            for i in range(len(tempp.child)):
                if (tempp.child[i] != None and tempp.child[i].key == value):
                    found = True;
 
            # If value is not the child of tempp node
            # then insert all the children
            # of the tempp node
            if (found == False):
                for i in range(len(tempp.child)):
                    if (tempp.child[i] != None):
                        q.append(tempp.child[i]);
 
            qsize -= 1
 
    if (found):
 
        # Queue will contain the cousins
        qsize = len(q);
 
        if (qsize == 0):
            print("NA");
        for i in range(qsize):
            tempp = q[0];
            q.pop(0);
            print(tempp.key, end= " ");
    else:
 
        # When value  is not in the tree
        print("Not Possible");
    print('')
    return;
 
# Driver Code
root = Node(10);
root.child.append(Node(77));
root.child.append(Node(90));
root.child.append(Node(35));
root.child.append(Node(19));
root.child[0].child.append(Node(88));
root.child[0].child.append(Node(98));
root.child[0].child[1].child.append(Node(76));
root.child[0].child[1].child.append(Node(20));
root.child[1].child.append(Node(61));
root.child[1].child.append(Node(74));
root.child[2].child.append(Node(39));
root.child[3].child.append(Node(17));
root.child[3].child.append(Node(72));
root.child[3].child.append(Node(19));
 
# Find the cousins of value
value = 39;
printCousins(root, value);
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
 
  // Structure of a node of N-ary tree
  class Node {
    public int key;
    public List<Node> child = new List<Node>();
  };
 
  // New node creation
  static Node newNode(int key)
  {
    Node temp = new Node();
    temp.key = key;
    return temp;
  }
 
  // Function to find the cousins of a
  // given node in an N-array tree
  static void printCousins(Node root, int value)
  {
 
    // Base case
    if (root == null)
      return;
 
    Queue<Node> q = new Queue<Node>();
    q.Enqueue(root);
 
    // If we find the node
    // with value as the key
    bool found = false;
 
    int qsize = 0;
    Node tempp;
 
    while (q.Count!=0 && !found) {
      qsize = q.Count;
 
      while (qsize > 0) {
 
        // Storing the current node
        tempp = q.Peek();
        q.Dequeue();
 
        // If we have already found
        // the value as child of a node,
        // we need to insert children of other
        // node of same level in the queue
        if (found == true) {
          for (int i = 0; i < tempp.child.Count;
               i++) {
            if (tempp.child[i] != null)
              q.Enqueue(tempp.child[i]);
          }
        }
 
        // If value is child of tempp node
        for (int i = 0; i < tempp.child.Count; i++)
          if (tempp.child[i] != null
              && tempp.child[i].key == value)
            found = true;
 
        // If value is not the child of tempp node
        // then insert all the children
        // of the tempp node
        if (found == false) {
          for (int i = 0; i < tempp.child.Count;
               i++) {
            if (tempp.child[i] != null)
              q.Enqueue(tempp.child[i]);
          }
        }
 
        qsize--;
      }
    }
 
    if (found) {
 
      // Queue will contain the cousins
      qsize = q.Count;
 
      if (qsize == 0)
        Console.Write("NA");
      for (int i = 0; i < qsize; i++) {
        tempp = q.Peek();
        q.Dequeue();
        Console.Write(tempp.key+ " ");
      }
    }
    else {
 
      // When value  is not in the tree
      Console.Write("Not Possible");
    }
    Console.Write("\n");
    return;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
 
    Node root = newNode(10);
    (root.child).Add(newNode(77));
    (root.child).Add(newNode(90));
    (root.child).Add(newNode(35));
    (root.child).Add(newNode(19));
    (root.child[0].child).Add(newNode(88));
    (root.child[0].child).Add(newNode(98));
    (root.child[0].child[1].child)
    .Add(newNode(76));
    (root.child[0].child[1].child)
    .Add(newNode(20));
    (root.child[1].child).Add(newNode(61));
    (root.child[1].child).Add(newNode(74));
    (root.child[2].child).Add(newNode(39));
    (root.child[3].child).Add(newNode(17));
    (root.child[3].child).Add(newNode(72));
    (root.child[3].child).Add(newNode(19));
 
    // Find the cousins of value
    int value = 39;
    printCousins(root, value);
 
  }
}
 
 
 
// This code is contributed by 29AjayKumar

Javascript

<script>
      // JavaScript code for the above approach 
 
      // Structure of a node of N-ary tree
      class Node {
          constructor(k) {
              this.key = k;
              this.child = [];
          }
      };
 
      // New node creation
 
      // Function to find the cousins of a
      // given node in an N-array tree
      function printCousins(root, value) {
 
          // Base case
          if (root == null)
              return;
 
          let q = [];
          q.push(root);
 
          // If we find the node
          // with value as the key
          let found = false;
 
          let qsize = 0;
          let tempp;
 
          while (q.length != 0 && found != 1) {
              qsize = q.length;
 
              while (qsize != 0) {
 
                  // Storing the current node
                  tempp = q[0];
                  q.shift();
 
                  // If we have already found
                  // the value as child of a node,
                  // we need to insert children of other
                  // node of same level in the queue
                  if (found == true) {
                      for (let i = 0; i < tempp.child.length;
                          i++) {
                          if (tempp.child[i] != null)
                              q.push(tempp.child[i]);
                      }
                  }
 
                  // If value is child of tempp node
                  for (let i = 0; i < tempp.child.length; i++)
                      if (tempp.child[i] != null
                          && tempp.child[i].key == value)
                          found = true;
 
                  // If value is not the child of tempp node
                  // then insert all the children
                  // of the tempp node
                  if (found == false) {
                      for (let i = 0; i < tempp.child.length;
                          i++) {
                          if (tempp.child[i] != null)
                              q.push(tempp.child[i]);
                      }
                  }
 
                  qsize--;
              }
          }
 
          if (found) {
 
              // Queue will contain the cousins
              qsize = q.length;
 
              if (qsize == 0)
                  document.write("NA");
              for (let i = 0; i < qsize; i++) {
                  tempp = q[0];
                  q.shift();
                  document.write(tempp.key + " ");
              }
          }
          else {
 
              // When value  is not in the tree
              document.write("Not Possible");
          }
          document.write('<br>')
          return;
      }
 
      // Driver Code
      let root = new Node(10);
      root.child.push(new Node(77));
      root.child.push(new Node(90));
      root.child.push(new Node(35));
      root.child.push(new Node(19));
      root.child[0].child.push(new Node(88));
      root.child[0].child.push(new Node(98));
      root.child[0].child[1].child
          .push(new Node(76));
      root.child[0].child[1].child
          .push(new Node(20));
      root.child[1].child.push(new Node(61));
      root.child[1].child.push(new Node(74));
      root.child[2].child.push(new Node(39));
      root.child[3].child.push(new Node(17));
      root.child[3].child.push(new Node(72));
      root.child[3].child.push(new Node(19));
 
      // Find the cousins of value
      let value = 39;
      printCousins(root, value);
 
// This code is contributed by Potta Lokesh
  </script>

Output

Cousins for the element 39: 88 98 61 74 17 72 19

Time Complexity: O(N)
Auxiliary Space: O(N)

Suggest improvement

Number of ways to traverse an N-ary tree

Height of n-ary tree if parent array is given

Share your thoughts in the comments

N-ary Tree Traversals

Easy problems on n-ary Tree

Intermediate problems on n-ary Tree

Hard problems on n-ary Tree