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Diameter of an N-ary tree
• Difficulty Level : Medium
• Last Updated : 13 Jan, 2021

The diameter of an N-ary tree is the longest path present between any two nodes of the tree. These two nodes must be two leaf nodes. The following examples have the longest path[diameter] shaded.

Example 1:

Example 2:

Prerequisite: Diameter of a binary tree.

The path can either start from one of the nodes and goes up to one of the LCAs of these nodes and again come down to the deepest node of some other subtree or can exist as a diameter of one of the child of the current node.
The solution will exist in any one of these:
I] Diameter of one of the children of the current node
II] Sum of Height of the highest two subtree + 1

## C++

 `// C++ program to find the height of an N-ary` `// tree` `#include ` `using` `namespace` `std;`   `// Structure of a node of an n-ary tree` `struct` `Node` `{` `    ``char` `key;` `    ``vector child;` `};`   `// Utility function to create a new tree node` `Node *newNode(``int` `key)` `{` `    ``Node *temp = ``new` `Node;` `    ``temp->key = key;` `    ``return` `temp;` `}`   `// Utility function that will return the depth` `// of the tree` `int` `depthOfTree(``struct` `Node *ptr)` `{` `    ``// Base case` `    ``if` `(!ptr)` `        ``return` `0;`   `    ``int` `maxdepth = 0;`   `    ``// Check for all children and find` `    ``// the maximum depth` `    ``for` `(vector::iterator it = ptr->child.begin();` `                           ``it != ptr->child.end(); it++)`   `        ``maxdepth = max(maxdepth , depthOfTree(*it));`   `    ``return` `maxdepth + 1;` `}`   `// Function to calculate the diameter` `// of the tree` `int` `diameter(``struct` `Node *ptr)` `{` `    ``// Base case` `    ``if` `(!ptr)` `        ``return` `0;`   `    ``// Find top two highest children` `    ``int` `max1 = 0, max2 = 0;` `    ``for` `(vector::iterator it = ptr->child.begin();` `                          ``it != ptr->child.end(); it++)` `    ``{` `        ``int` `h = depthOfTree(*it);` `        ``if` `(h > max1)` `           ``max2 = max1, max1 = h;` `        ``else` `if` `(h > max2)` `           ``max2 = h;` `    ``}`   `    ``// Iterate over each child for diameter` `    ``int` `maxChildDia = 0;` `    ``for` `(vector::iterator it = ptr->child.begin();` `                           ``it != ptr->child.end(); it++)` `        ``maxChildDia = max(maxChildDia, diameter(*it));`   `    ``return` `max(maxChildDia, max1 + max2 + 1);` `}`   `// Driver program` `int` `main()` `{` `    ``/*   Let us create below tree` `    ``*           A` `    ``*         / /  \  \` `    ``*       B  F   D  E` `    ``*      / \     |  /|\` `    ``*     K  J    G  C H I` `    ``*      /\            \` `    ``*    N   M            L` `    ``*/`   `    ``Node *root = newNode(``'A'``);` `    ``(root->child).push_back(newNode(``'B'``));` `    ``(root->child).push_back(newNode(``'F'``));` `    ``(root->child).push_back(newNode(``'D'``));` `    ``(root->child).push_back(newNode(``'E'``));` `    ``(root->child[0]->child).push_back(newNode(``'K'``));` `    ``(root->child[0]->child).push_back(newNode(``'J'``));` `    ``(root->child[2]->child).push_back(newNode(``'G'``));` `    ``(root->child[3]->child).push_back(newNode(``'C'``));` `    ``(root->child[3]->child).push_back(newNode(``'H'``));` `    ``(root->child[3]->child).push_back(newNode(``'I'``));` `    ``(root->child[0]->child[0]->child).push_back(newNode(``'N'``));` `    ``(root->child[0]->child[0]->child).push_back(newNode(``'M'``));` `    ``(root->child[3]->child[2]->child).push_back(newNode(``'L'``));`   `    ``cout << diameter(root) << endl;`   `    ``return` `0;` `}`

## Java

 `// Java program to find the height of an N-ary` `// tree` `import` `java.util.*;` `class` `GFG` `{`   `// Structure of a node of an n-ary tree` `static` `class` `Node` `{` `    ``char` `key;` `    ``Vector child;` `};`   `// Utility function to create a new tree node` `static` `Node newNode(``int` `key)` `{` `    ``Node temp = ``new` `Node();` `    ``temp.key = (``char``) key;` `    ``temp.child = ``new` `Vector();` `    ``return` `temp;` `}`   `// Utility function that will return the depth` `// of the tree` `static` `int` `depthOfTree(Node ptr)` `{` `    ``// Base case` `    ``if` `(ptr == ``null``)` `        ``return` `0``;`   `    ``int` `maxdepth = ``0``;`   `    ``// Check for all children and find` `    ``// the maximum depth` `    ``for` `(Node it : ptr.child)`   `        ``maxdepth = Math.max(maxdepth,` `                            ``depthOfTree(it));`   `    ``return` `maxdepth + ``1``;` `}`   `// Function to calculate the diameter` `// of the tree` `static` `int` `diameter(Node ptr)` `{` `    ``// Base case` `    ``if` `(ptr == ``null``)` `        ``return` `0``;`   `    ``// Find top two highest children` `    ``int` `max1 = ``0``, max2 = ``0``;` `    ``for` `(Node it : ptr.child)` `    ``{` `        ``int` `h = depthOfTree(it);` `        ``if` `(h > max1)` `        ``{` `            ``max2 = max1;` `            ``max1 = h;` `        ``}` `        ``else` `if` `(h > max2)` `        ``max2 = h;` `    ``}`   `    ``// Iterate over each child for diameter` `    ``int` `maxChildDia = ``0``;` `    ``for` `(Node it : ptr.child)` `        ``maxChildDia = Math.max(maxChildDia, ` `                               ``diameter(it));`   `    ``return` `Math.max(maxChildDia, max1 + max2 + ``1``);` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``/* Let us create below tree` `    ``*         A` `    ``*         / / \ \` `    ``*     B F D E` `    ``*     / \     | /|\` `    ``*     K J G C H I` `    ``*     /\         \` `    ``* N M         L` `    ``*/` `    ``Node root = newNode(``'A'``);` `    ``(root.child).add(newNode(``'B'``));` `    ``(root.child).add(newNode(``'F'``));` `    ``(root.child).add(newNode(``'D'``));` `    ``(root.child).add(newNode(``'E'``));` `    ``(root.child.get(``0``).child).add(newNode(``'K'``));` `    ``(root.child.get(``0``).child).add(newNode(``'J'``));` `    ``(root.child.get(``2``).child).add(newNode(``'G'``));` `    ``(root.child.get(``3``).child).add(newNode(``'C'``));` `    ``(root.child.get(``3``).child).add(newNode(``'H'``));` `    ``(root.child.get(``3``).child).add(newNode(``'I'``));` `    ``(root.child.get(``0``).child.get(``0``).child).add(newNode(``'N'``));` `    ``(root.child.get(``0``).child.get(``0``).child).add(newNode(``'M'``));` `    ``(root.child.get(``3``).child.get(``2``).child).add(newNode(``'L'``));`   `    ``System.out.print(diameter(root) + ``"\n"``);` `}` `}`   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python program to find the height of an N-ary` `# tree`   `# Structure of a node of an n-ary tree` `class` `Node:` `    ``def` `__init__(``self``, x):` `        ``self``.key ``=` `x` `        ``self``.child ``=` `[]`   `# Utility function that will return the depth` `# of the tree` `def` `depthOfTree(ptr):` `    `  `    ``# Base case` `    ``if` `(``not` `ptr):` `        ``return` `0` `    ``maxdepth ``=` `0`   `    ``# Check for all children and find` `    ``# the maximum depth` `    ``for` `it ``in` `ptr.child:` `        ``maxdepth ``=` `max``(maxdepth , depthOfTree(it))` `    ``return` `maxdepth ``+` `1`   `# Function to calculate the diameter` `# of the tree` `def` `diameter(ptr):` `    `  `    ``# Base case` `    ``if` `(``not` `ptr):` `        ``return` `0`   `    ``# Find top two highest children` `    ``max1, max2 ``=` `0``, ``0` `    ``for` `it ``in` `ptr.child:` `        ``h ``=` `depthOfTree(it)` `        ``if` `(h > max1):` `           ``max2, max1 ``=` `max1, h` `        ``elif` `(h > max2):` `           ``max2 ``=` `h`   `    ``# Iterate over each child for diameter` `    ``maxChildDia ``=` `0` `    ``for` `it ``in` `ptr.child:` `        ``maxChildDia ``=` `max``(maxChildDia, diameter(it))` `    ``return` `max``(maxChildDia, max1 ``+` `max2 ``+` `1``)`   `# Driver program` `if` `__name__ ``=``=` `'__main__'``:` `    ``# /*   Let us create below tree` `    ``# *           A` `    ``# *         / /  \  \` `    ``# *       B  F   D  E` `    ``# *      / \     |  /|\` `    ``# *     K  J    G  C H I` `    ``# *      /\            \` `    ``# *    N   M            L` `    ``# */`   `    ``root ``=` `Node(``'A'``)` `    ``(root.child).append(Node(``'B'``))` `    ``(root.child).append(Node(``'F'``))` `    ``(root.child).append(Node(``'D'``))` `    ``(root.child).append(Node(``'E'``))` `    ``(root.child[``0``].child).append(Node(``'K'``))` `    ``(root.child[``0``].child).append(Node(``'J'``))` `    ``(root.child[``2``].child).append(Node(``'G'``))` `    ``(root.child[``3``].child).append(Node(``'C'``))` `    ``(root.child[``3``].child).append(Node(``'H'``))` `    ``(root.child[``3``].child).append(Node(``'I'``))` `    ``(root.child[``0``].child[``0``].child).append(Node(``'N'``))` `    ``(root.child[``0``].child[``0``].child).append(Node(``'M'``))` `    ``(root.child[``3``].child[``2``].child).append(Node(``'L'``))`   `    ``print``(diameter(root))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to find the height of ` `// an N-ary tree` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG` `{`   `// Structure of a node of an n-ary tree` `class` `Node` `{` `    ``public` `char` `key;` `    ``public` `List child;` `};`   `// Utility function to create ` `// a new tree node` `static` `Node newNode(``int` `key)` `{` `    ``Node temp = ``new` `Node();` `    ``temp.key = (``char``) key;` `    ``temp.child = ``new` `List();` `    ``return` `temp;` `}`   `// Utility function that will return ` `// the depth of the tree` `static` `int` `depthOfTree(Node ptr)` `{` `    ``// Base case` `    ``if` `(ptr == ``null``)` `        ``return` `0;`   `    ``int` `maxdepth = 0;`   `    ``// Check for all children and find` `    ``// the maximum depth` `    ``foreach` `(Node it ``in` `ptr.child)`   `        ``maxdepth = Math.Max(maxdepth,` `                            ``depthOfTree(it));`   `    ``return` `maxdepth + 1;` `}`   `// Function to calculate the diameter` `// of the tree` `static` `int` `diameter(Node ptr)` `{` `    ``// Base case` `    ``if` `(ptr == ``null``)` `        ``return` `0;`   `    ``// Find top two highest children` `    ``int` `max1 = 0, max2 = 0;` `    ``foreach` `(Node it ``in` `ptr.child)` `    ``{` `        ``int` `h = depthOfTree(it);` `        ``if` `(h > max1)` `        ``{` `            ``max2 = max1;` `            ``max1 = h;` `        ``}` `        ``else` `if` `(h > max2)` `        ``max2 = h;` `    ``}`   `    ``// Iterate over each child for diameter` `    ``int` `maxChildDia = 0;` `    ``foreach` `(Node it ``in` `ptr.child)` `        ``maxChildDia = Math.Max(maxChildDia, ` `                               ``diameter(it));`   `    ``return` `Math.Max(maxChildDia, ` `                    ``max1 + max2 + 1);` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``/* Let us create below tree` `    ``*         A` `    ``*         / / \ \` `    ``*     B F D E` `    ``*     / \     | /|\` `    ``*     K J G C H I` `    ``*     /\         \` `    ``* N M         L` `    ``*/` `    ``Node root = newNode(``'A'``);` `    ``(root.child).Add(newNode(``'B'``));` `    ``(root.child).Add(newNode(``'F'``));` `    ``(root.child).Add(newNode(``'D'``));` `    ``(root.child).Add(newNode(``'E'``));` `    ``(root.child[0].child).Add(newNode(``'K'``));` `    ``(root.child[0].child).Add(newNode(``'J'``));` `    ``(root.child[2].child).Add(newNode(``'G'``));` `    ``(root.child[3].child).Add(newNode(``'C'``));` `    ``(root.child[3].child).Add(newNode(``'H'``));` `    ``(root.child[3].child).Add(newNode(``'I'``));` `    ``(root.child[0].child[0].child).Add(newNode(``'N'``));` `    ``(root.child[0].child[0].child).Add(newNode(``'M'``));` `    ``(root.child[3].child[2].child).Add(newNode(``'L'``));`   `    ``Console.Write(diameter(root) + ``"\n"``);` `}` `}`   `// This code is contributed by Rajput-Ji`

Output

`7`

Optimizations to above solution :

We can make a hash table to store heights of all nodes. If we precompute these heights, we don’t need to call depthOfTree() for every node.

A different optimized solution: Longest path in an undirected tree

Another Approach to get diameter using DFS in one traversal:

The diameter of a tree can be calculated as for every node

• The current node isn’t part of diameter (i.e Diameter lies on of one of the children of the current node).
• The current node is part of diameter (i.e Diameter passes through the current node).

Node: Adjacency List has been used to store the Tree.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find` `// diameter of a tree using` `// DFS in ONE TRAVERSAL`   `#include ` `using` `namespace` `std;` `#define maxN 10005`   `// The array to store the` `// height of the nodes` `int` `height[maxN];`   `// Adjacency List to store` `// the tree` `vector<``int``> tree[maxN];`   `// varaiable to store diameter` `// of the tree` `int` `diameter = 0;`   `// Function to add edge between` `// node u to node v` `void` `addEdge(``int` `u, ``int` `v)` `{` `    ``// add edge from u to v` `    ``tree[u].push_back(v);`   `    ``// add edge from v to u` `    ``tree[v].push_back(u);` `}`   `void` `dfs(``int` `cur, ``int` `par)` `{` `    ``// Variables to store the height of children` `    ``// of cur node with maximum heights` `    ``int` `max1 = 0;` `    ``int` `max2 = 0;`   `    ``// going in the adjacency list of the current node` `    ``for` `(``auto` `u : tree[cur]) {` `        `  `        ``// if that node equals parent discard it` `        ``if` `(u == par)` `            ``continue``;`   `        ``// calling dfs for child node` `        ``dfs(u, cur);`   `        ``// calculating height of nodes` `        ``height[cur] = max(height[cur], height[u]);`   `        ``// getting the height of children` `        ``// of cur node with maximum height` `        ``if` `(height[u] >= max1) {` `            ``max2 = max1;` `            ``max1 = height[u];` `        ``}` `        ``else` `if` `(height[u] > max2) {` `            ``max2 = height[u];` `        ``}` `    ``}`   `    ``height[cur] += 1;`   `    ``// Diameter of a tree can be calculated as` `    ``// diameter passing through the node` `    ``// diameter doesn't includes the cur node` `    ``diameter = max(diameter, height[cur]);` `    ``diameter = max(diameter, max1 + max2 + 1);` `}`   `// Driver Code` `int` `main()` `{` `    ``// n is the number of nodes in tree` `    ``int` `n = 7;`   `    ``// Adding edges to the tree` `    ``addEdge(1, 2);` `    ``addEdge(1, 3);` `    ``addEdge(1, 4);` `    ``addEdge(2, 5);` `    ``addEdge(4, 6);` `    ``addEdge(4, 7);`   `    ``// Calling the dfs function to` `    ``// calculate the diameter of tree` `    ``dfs(1, 0);`   `    ``cout << ``"Diameter of tree is : "` `<< diameter - 1` `         ``<< ``"\n"``;`   `    ``return` `0;` `}`

Output

`Diameter of tree is : 4`

This article is contributed by Shubham Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.