# Diameter of n-ary tree using BFS

• Difficulty Level : Hard
• Last Updated : 14 Mar, 2023

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.

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

Implementation:

## C++

 `// C++ Program to find Diameter of n-ary tree``#include ``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 Code``int` `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;``}`

## Java

 `// Java Program to find Diameter of n-ary tree``import` `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>arr, ``int` `n)``{``    ``// Initializing queue``    ``Queue 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> arr, ``int` `n)``{``    ``int` `init = bfs(``1``, arr, n);``    ``int` `val = bfs(init, arr, n);``    ``return` `diameter[val];``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``// Input number of nodes``    ``int` `n = ``6``;` `    ``Vector> arr = ``new``    ``Vector>();``    ` `    ``for``(``int` `i = ``0``; i < n + ``1``; i++)``    ``{``        ``arr.add(``new` `Vector());``    ``}` `    ``// 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`

## 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` `ing` `def` `findDiameter(arr, n):` `    ``init ``=` `bfs(``1``, arr, n)``    ``val ``=` `bfs(init, arr, n)``    ``return` `diameter[val]` `# Driver Code``if` `__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`

## C#

 `// C# Program to find Diameter of n-ary tree``using` `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>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> arr, ``int` `n)``{``    ``int` `init = bfs(1, arr, n);``    ``int` `val = bfs(init, arr, n);``    ``return` `diameter[val];``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ``// Input number of nodes``    ``int` `n = 6;` `    ``List> arr = ``new``    ``List>();``    ` `    ``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`

## Javascript

 ``

Output

`Diameter of n-ary tree is 3`

Time complexity: O(n^2)
Auxiliary Space: O(10001)

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