Given a tree, with N nodes and E edges (every edge is denoted by two integers, X, Y stating that X is the parent of Y), the task is to print all the nodes with their right siblings in separate lines.
If there is no right sibling for a particular node then print -1 instead.
Examples:
In the image shown above, nodes 3, 5, 6, 7 are the right siblings of nodes 2, 4, 5, 6 respectively and the nodes 1, 3 and 7 have no right siblings.
Input: N = 7, E = 6
edges[][] = {{1, 2}, {1, 3}, {2, 4}, {2, 5}, {3, 6}, {3, 7}}
Output: 1 -1
2 3
3 -1
4 5
5 6
6 7
7 -1Input: N = 5, E = 4
edges[][] = {{7, 8}, {7, 10}, {7, 15}, {10, 3}}
Output: 7 -1
8 10
10 15
15 -1
3 -1
Approach: The main idea is to use a Breadth First Traversal.
- Initially, the root node and ‘-1’ value will be pushed into the queue. After every node from a particular level in the tree has been pushed to the queue, ‘-1’ has to be pushed to make sure that the last node in the level has no right sibling.
- After popping each node from the queue, the node at the front of the queue will always be the right sibling of the popped node.
- If the popped node is valued ‘-1’, it means the current level has been visited and if the queue is not empty it means that previous nodes of this level have at least one child which hasn’t been visited.
- Repeat the above steps while the queue is non-empty.
Below is the implementation of the above approach:
C++
// C++ program to print right siblings// of all the nodes in a tree#include <bits/stdc++.h>using namespace std;void PrintSiblings(int root, int N, int E, vector<int> adj[]){ // boolean array to mark the visited nodes vector<bool> vis(N+1, false); // queue data structure to implement bfs queue<int> q; q.push(root); q.push(-1); vis[root] = 1; while (!q.empty()) { int node = q.front(); q.pop(); if (node == -1) { // if queue is empty then // the popped node is the last node // no need to push -1. if (!q.empty()) q.push(-1); continue; } // node and its right sibling cout << node << " " << q.front() << "\n"; for (auto s : adj[node]) { if (!vis[s]) { vis[s] = 1; q.push(s); } } }}// Driver codeint main(){ // nodes and edges int N = 7, E = 6; vector<int> adj[N+1]; // The tree is represented in the form of // an adjacency list as there can be // multiple children of a node adj[1].push_back(2); adj[1].push_back(3); adj[2].push_back(4); adj[2].push_back(5); adj[3].push_back(6); adj[3].push_back(7); int root = 1; PrintSiblings(root, N, E, adj); return 0;} |
Java
// Java program to print right siblings// of all the nodes in a treeimport java.util.*;class GFG{static void PrintSiblings(int root, int N, int E, Vector<Integer> adj[]){ // boolean array to mark the visited nodes boolean []vis = new boolean[N + 1]; // queue data structure to implement bfs Queue<Integer> q = new LinkedList<>(); q.add(root); q.add(-1); vis[root] = true; while (!q.isEmpty()) { int node = q.peek(); q.remove(); if (node == -1) { // if queue is empty then // the popped node is the last node // no need to push -1. if (!q.isEmpty()) q.add(-1); continue; } // node and its right sibling System.out.print(node + " " + q.peek() + "\n"); for (Integer s : adj[node]) { if (!vis[s]) { vis[s] = true; q.add(s); } } }}// Driver codepublic static void main(String[] args){ // nodes and edges int N = 7, E = 6; Vector<Integer> []adj = new Vector[N + 1]; for(int i = 0; i < N + 1; i++) adj[i] = new Vector<Integer>(); // The tree is represented in the form of // an adjacency list as there can be // multiple children of a node adj[1].add(2); adj[1].add(3); adj[2].add(4); adj[2].add(5); adj[3].add(6); adj[3].add(7); int root = 1; PrintSiblings(root, N, E, adj);}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to print right# siblings of all the nodes in # a treedef PrintSiblings(root, N, E, adj): # Boolean array to mark the # visited nodes vis = [False for i in range(N + 1)] # queue data structure to # implement bfs q = [] q.append(root) q.append(-1) vis[root] = 1 while (len(q) != 0): node = q[0] q.pop(0) if (node == -1): # If queue is empty then the # popped node is the last node # no need to append -1. if (len(q) != 0): q.append(-1) continue # Node and its right sibling print(str(node) + " " + str(q[0])) for s in adj[node]: if (not vis[s]): vis[s] = True q.append(s) # Driver codeif __name__=='__main__': # Nodes and edges N = 7 E = 6 adj = [[] for i in range(N + 1)] # The tree is represented in the # form of an adjacency list as # there can be multiple children # of a node adj[1].append(2) adj[1].append(3) adj[2].append(4) adj[2].append(5) adj[3].append(6) adj[3].append(7) root = 1 PrintSiblings(root, N, E, adj) # This code is contributed by rutvik_56 |
C#
// C# program to print right siblings// of all the nodes in a treeusing System;using System.Collections.Generic;class GFG{static void PrintSiblings(int root, int N, int E, List<int> []adj){ // bool array to mark the visited nodes bool []vis = new bool[N + 1]; // queue data structure to implement bfs Queue<int> q = new Queue<int>(); q.Enqueue(root); q.Enqueue(-1); vis[root] = true; while (q.Count != 0) { int node = q.Peek(); q.Dequeue(); if (node == -1) { // if queue is empty then // the popped node is the last node // no need to push -1. if (q.Count != 0) q.Enqueue(-1); continue; } // node and its right sibling Console.Write(node + " " + q.Peek() + "\n"); foreach (int s in adj[node]) { if (!vis[s]) { vis[s] = true; q.Enqueue(s); } } }}// Driver codepublic static void Main(String[] args){ // nodes and edges int N = 7, E = 6; List<int> []adj = new List<int>[N + 1]; for(int i = 0; i < N + 1; i++) adj[i] = new List<int>(); // The tree is represented in the form of // an adjacency list as there can be // multiple children of a node adj[1].Add(2); adj[1].Add(3); adj[2].Add(4); adj[2].Add(5); adj[3].Add(6); adj[3].Add(7); int root = 1; PrintSiblings(root, N, E, adj);}}// This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program to print right siblings // of all the nodes in a tree function PrintSiblings(root, N, E, adj) { // boolean array to mark the visited nodes let vis = new Array(N + 1); // queue data structure to implement bfs let q = []; q.push(root); q.push(-1); vis[root] = true; while (q.length > 0) { let node = q[0]; q.shift(); if (node == -1) { // if queue is empty then // the popped node is the last node // no need to push -1. if (q.length > 0) q.push(-1); continue; } // node and its right sibling document.write(node + " " + q[0] + "</br>"); for (let s = 0; s < adj[node].length; s++) { if (!vis[adj[node][s]]) { vis[adj[node][s]] = true; q.push(adj[node][s]); } } } } // nodes and edges let N = 7, E = 6; let adj = new Array(N + 1); for(let i = 0; i < N + 1; i++) adj[i] = []; // The tree is represented in the form of // an adjacency list as there can be // multiple children of a node adj[1].push(2); adj[1].push(3); adj[2].push(4); adj[2].push(5); adj[3].push(6); adj[3].push(7); let root = 1; PrintSiblings(root, N, E, adj);// This code is contributed by decode2207.</script> |
Output
1 -1 2 3 3 -1 4 5 5 6 6 7 7 -1
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
