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 19Input: K = 17
Output: 88 98 61 74 39Input: 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 treestruct Node { int key; vector<Node*> child;};// New node creationNode* 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 treevoid 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 Codeint 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 approachimport 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 treeclass 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 treedef 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 Coderoot = 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 valuevalue = 39;printCousins(root, value);# This code is contributed by gfgking |
C#
// C# program for the above approachusing 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)
