Diameter of n-ary tree using BFS
N-ary tree refers to the rooted tree in which each node having atmost k child nodes. The diameter of n-ary tree is the longest path between two leaf nodes.
Various approaches have already been discussed to compute diameter of tree.
1. Diameter of an N-ary tree
2. Diameter of a Binary Tree in O(n)
3. Diameter of a Binary Tree
4. Diameter of a tree using DFS
This article discuss another approach for computing diameter tree of n-ary tree using bfs.
Step 1: Run bfs to find the farthest node from rooted tree let say A
Step 2: Then run bfs from A to find farthest node from A let B
Step 3: Distance between node A and B is the diameter of given tree
C++
// C++ Program to find Diameter of n-ary tree#include <bits/stdc++.h>using namespace std;// Here 10000 is maximum number of nodes in// given tree.int diameter[10001];// The Function to do bfs traversal.// It uses iterative approach to do bfs// bfsUtil()int bfs(int init, vector<int> arr[], int n){ // Initializing queue queue<int> q; q.push(init); int visited[n + 1]; for (int i = 0; i <= n; i++) { visited[i] = 0; diameter[i] = 0; } // Pushing each node in queue q.push(init); // Mark the traversed node visited visited[init] = 1; while (!q.empty()) { int u = q.front(); q.pop(); for (int i = 0; i < arr[u].size(); i++) { if (visited[arr[u][i]] == 0) { visited[arr[u][i]] = 1; // Considering weight of edges equal to 1 diameter[arr[u][i]] += diameter[u] + 1; q.push(arr[u][i]); } } } // return index of max value in diameter return int(max_element(diameter + 1, diameter + n + 1) - diameter);}int findDiameter(vector<int> arr[], int n){ int init = bfs(1, arr, n); int val = bfs(init, arr, n); return diameter[val];}// Driver Codeint main(){ // Input number of nodes int n = 6; vector<int> arr[n + 1]; // Input nodes in adjacency list arr[1].push_back(2); arr[1].push_back(3); arr[1].push_back(6); arr[2].push_back(4); arr[2].push_back(1); arr[2].push_back(5); arr[3].push_back(1); arr[4].push_back(2); arr[5].push_back(2); arr[6].push_back(1); printf("Diameter of n-ary tree is %d\n", findDiameter(arr, n)); return 0;} |
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Java
// Java Program to find Diameter of n-ary treeimport java.util.*;class GFG{ // Here 10000 is maximum number of nodes in// given tree.static int diameter[] = new int[10001];// The Function to do bfs traversal.// It uses iterative approach to do bfs// bfsUtil()static int bfs(int init, Vector<Vector<Integer>>arr, int n){ // Initializing queue Queue<Integer> q = new LinkedList<>(); q.add(init); int visited[] = new int[n + 1]; for (int i = 0; i <= n; i++) { visited[i] = 0; diameter[i] = 0; } // Pushing each node in queue q.add(init); // Mark the traversed node visited visited[init] = 1; while (q.size() > 0) { int u = q.peek(); q.remove(); for (int i = 0; i < arr.get(u).size(); i++) { if (visited[arr.get(u).get(i)] == 0) { visited[arr.get(u).get(i)] = 1; // Considering weight of edges equal to 1 diameter[arr.get(u).get(i)] += diameter[u] + 1; q.add(arr.get(u).get(i)); } } } int in = 0; for(int i = 0; i <= n; i++) { if(diameter[i] > diameter[in]) in = i; } // return index of max value in diameter return in;}static int findDiameter(Vector<Vector<Integer>> arr, int n){ int init = bfs(1, arr, n); int val = bfs(init, arr, n); return diameter[val];}// Driver Codepublic static void main(String args[]){ // Input number of nodes int n = 6; Vector<Vector<Integer>> arr = new Vector<Vector<Integer>>(); for(int i = 0; i < n + 1; i++) { arr.add(new Vector<Integer>()); } // Input nodes in adjacency list arr.get(1).add(2); arr.get(1).add(3); arr.get(1).add(6); arr.get(2).add(4); arr.get(2).add(1); arr.get(2).add(5); arr.get(3).add(1); arr.get(4).add(2); arr.get(5).add(2); arr.get(6).add(1); System.out.printf("Diameter of n-ary tree is %d\n", findDiameter(arr, n));}}// This code is contributed by Arnab Kundu |
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Python3
# Python3 program to find diameter of n-ary tree # Here 10000 is maximum number of nodes in# given tree.diameter = [0 for i in range(10001)] # The Function to do bfs traversal.# It uses iterative approach to do bfs# bfsUtil()def bfs(init, arr, n): # Initializing queue q = [] q.append(init) visited = [0 for i in range(n + 1)] for i in range(n + 1): visited[i] = 0 diameter[i] = 0 # Pushing each node in queue q.append(init) # Mark the traversed node visited visited[init] = 1 while (len(q) > 0): u = q[0] q.pop(0) for i in range(len(arr[u])): if (visited[arr[u][i]] == 0): visited[arr[u][i]] = 1 # Considering weight of edges equal to 1 diameter[arr[u][i]] += diameter[u] + 1 q.append(arr[u][i]) ing = 0 for i in range(n + 1): if(diameter[i] > diameter[ing]): ing = i # Return index of max value in diameter return ingdef findDiameter(arr, n): init = bfs(1, arr, n) val = bfs(init, arr, n) return diameter[val]# Driver Codeif __name__=='__main__': # Input number of nodes n = 6 arr = [[] for i in range(n + 1)] # Input nodes in adjacency list arr[1].append(2) arr[1].append(3) arr[1].append(6) arr[2].append(4) arr[2].append(1) arr[2].append(5) arr[3].append(1) arr[4].append(2) arr[5].append(2) arr[6].append(1) print("Diameter of n-ary tree is " + str(findDiameter(arr, n)))# This code is contributed by rutvik_56 |
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C#
// C# Program to find Diameter of n-ary treeusing System;using System.Collections.Generic;class GFG{ // Here 10000 is maximum number of nodes // in given tree.static int []diameter = new int[10001];// The Function to do bfs traversal.// It uses iterative approach to do bfs// bfsUtil()static int bfs(int init, List<List<int>>arr, int n){ // Initializing queue Queue<int> q = new Queue<int>(); q.Enqueue(init); int []visited = new int[n + 1]; for (int i = 0; i <= n; i++) { visited[i] = 0; diameter[i] = 0; } // Pushing each node in queue q.Enqueue(init); // Mark the traversed node visited visited[init] = 1; while (q.Count > 0) { int u = q.Peek(); q.Dequeue(); for (int i = 0; i < arr[u].Count; i++) { if (visited[arr[u][i]] == 0) { visited[arr[u][i]] = 1; // Considering weight of edges equal to 1 diameter[arr[u][i]] += diameter[u] + 1; q.Enqueue(arr[u][i]); } } } int iN = 0; for(int i = 0; i <= n; i++) { if(diameter[i] > diameter[iN]) iN = i; } // return index of max value in diameter return iN;}static int findDiameter(List<List<int>> arr, int n){ int init = bfs(1, arr, n); int val = bfs(init, arr, n); return diameter[val];}// Driver Codepublic static void Main(String []args){ // Input number of nodes int n = 6; List<List<int>> arr = new List<List<int>>(); for(int i = 0; i < n + 1; i++) { arr.Add(new List<int>()); } // Input nodes in adjacency list arr[1].Add(2); arr[1].Add(3); arr[1].Add(6); arr[2].Add(4); arr[2].Add(1); arr[2].Add(5); arr[3].Add(1); arr[4].Add(2); arr[5].Add(2); arr[6].Add(1); Console.Write("Diameter of n-ary tree is {0}\n", findDiameter(arr, n));}}// This code is contributed by PrinciRaj1992 |
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Output:
Diameter of n-ary tree is 3
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