Diameter of an N-ary tree - GeeksforGeeks

31 Aug, 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 go 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++
Java
Python3
C#
Javascript
// C++ program to find the height of an N-ary// tree#include <bits/stdc++.h>using namespace std; // Structure of a node of an n-ary treestruct Node{    char key;    vector<Node *> child;}; // Utility function to create a new tree nodeNode *newNode(int key){    Node *temp = new Node;    temp->key = key;    return temp;} // Utility function that will return the depth// of the treeint depthOfTree(struct Node *ptr){    // Base case    if (!ptr)        return 0;     int maxdepth = 0;     // Check for all children and find    // the maximum depth    for (vector<Node*>::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 treeint diameter(struct Node *ptr){    // Base case    if (!ptr)        return 0;     // Find top two highest children    int max1 = 0, max2 = 0;    for (vector<Node*>::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<Node*>::iterator it = ptr->child.begin();                           it != ptr->child.end(); it++)        maxChildDia = max(maxChildDia, diameter(*it));     return max(maxChildDia, max1 + max2 + 1);} // Driver programint 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 program to find the height of an N-ary// treeimport java.util.*;class GFG{ // Structure of a node of an n-ary treestatic class Node{    char key;    Vector<Node> child;}; // Utility function to create a new tree nodestatic Node newNode(int key){    Node temp = new Node();    temp.key = (char) key;    temp.child = new Vector<Node>();    return temp;} // Utility function that will return the depth// of the treestatic 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 treestatic 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 Codepublic 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
# Python program to find the height of an N-ary# tree # Structure of a node of an n-ary treeclass Node:    def __init__(self, x):        self.key = x        self.child = [] # Utility function that will return the depth# of the treedef 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 treedef 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 programif __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# program to find the height of// an N-ary treeusing System;using System.Collections.Generic; class GFG{ // Structure of a node of an n-ary treeclass Node{    public char key;    public List<Node> child;}; // Utility function to create// a new tree nodestatic Node newNode(int key){    Node temp = new Node();    temp.key = (char) key;    temp.child = new List<Node>();    return temp;} // Utility function that will return// the depth of the treestatic 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 treestatic 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 Codepublic 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
<script> // Javascript program to find the// height of an N-ary tree         // Structure of a node of an n-ary tree    class Node{                 // Utility function to create a new tree node        constructor(key)        {            this.key=key;            this.child=[];        }    }         // Utility function that will    // return the depth    // of the tree    function depthOfTree(ptr)    {        // Base case    if (ptr == null)        return 0;      let maxdepth = 0;      // Check for all children and find    // the maximum depth    for (let it=0;it< ptr.child.length;it++)          maxdepth = Math.max(maxdepth,                     depthOfTree(ptr.child[it]));      return maxdepth + 1;    }          // Function to calculate the diameter// of the tree    function diameter(ptr)    {        // Base case    if (ptr == null)        return 0;      // Find top two highest children    let max1 = 0, max2 = 0;    for (let it=0;it< ptr.child.length;it++)    {        let h = depthOfTree(ptr.child[it]);        if (h > max1)        {            max2 = max1;            max1 = h;        }        else if (h > max2)        max2 = h;    }      // Iterate over each child for diameter    let maxChildDia = 0;    for (let it=0;it< ptr.child.length;it++)        maxChildDia = Math.max(maxChildDia,                          diameter(ptr.child[it]));      return Math.max(maxChildDia, max1 + max2 + 1);    }         // Driver Code         /* Let us create below tree    *         A    *         / / \ \    *     B F D E    *     / \     | /|\    *     K J G C H I    *     /\         \    * N M         L    */    let root = new Node('A');    (root.child).push(new Node('B'));    (root.child).push(new Node('F'));    (root.child).push(new Node('D'));    (root.child).push(new Node('E'));    (root.child[0].child).push(new Node('K'));    (root.child[0].child).push(new Node('J'));    (root.child[2].child).push(new Node('G'));    (root.child[3].child).push(new Node('C'));    (root.child[3].child).push(new Node('H'));    (root.child[3].child).push(new Node('I'));    (root.child[0].child[0].child).push(new Node('N'));    (root.child[0].child[0].child).push(new Node('M'));    (root.child[3].child[2].child).push(new Node('L'));      document.write(diameter(root) + "\n"); // This code is contributed by patel2127 </script>
7
Optimizations to above solution :  
We can find diameter without calculating depth of the tree making small changes in the above solution, similar to finding diameter of binary tree.
C++
Java
// C++ program to find the height of an N-ary// tree#include <bits/stdc++.h>using namespace std; // Structure of a node of an n-ary treestruct Node{    char key;    vector<Node *> child;}; // Utility function to create a new tree nodeNode *newNode(int key){    Node *temp = new Node;    temp->key = key;    return temp;} int diameter(struct Node *ptr,int &diameter_of_tree){    // Base case    if (!ptr)        return 0;     // Find top two highest children    int max1 = 0, max2 = 0;    for (vector<Node*>::iterator it = ptr->child.begin();it != ptr->child.end(); it++)    {        int h = diameter(*it,diameter_of_tree);        if (h > max1)        max2 = max1, max1 = h;        else if (h > max2)        max2 = h;    }     // Find whether our node can be part of diameter    diameter_of_tree = max(max1 + max2 + 1,diameter_of_tree);     return max(max1,max2) + 1;} 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'));         // for storing diameter    int diameter_of_tree = 0;         diameter(root,diameter_of_tree);         cout << diameter_of_tree << endl;     return 0;}// This code is improved by bhuvan
// Java program to find the height of an N-ary// treeimport java.util.*;class GFG {     // Structure of a node of an n-ary tree    static class Node {        char key;        Vector<Node> 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<Node>();        return temp;    }     // for storing diameter_of_tree    public static int diameter_of_tree = 0;     // 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 = diameter(it);            if (h > max1) {                max2 = max1;                max1 = h;            }            else if (h > max2)                max2 = h;        }         diameter_of_tree            = Math.max(max1 + max2 + 1, diameter_of_tree);        return (Math.max(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'));         diameter(root);        System.out.print(diameter_of_tree + "\n");    }} // This code is improved by Bhuvan
Output
7
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 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++
Java
Python3
Javascript
// C++ implementation to find// diameter of a tree using// DFS in ONE TRAVERSAL #include <bits/stdc++.h>using namespace std;#define maxN 10005 // The array to store the// height of the nodesint height[maxN]; // Adjacency List to store// the treevector<int> tree[maxN]; // variable to store diameter// of the treeint diameter = 0; // Function to add edge between// node u to node vvoid 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 Codeint 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;}
/*package whatever //do not write package name here */import java.io.*;import java.util.*; class GFG {   static int maxN = 10005;   // The array to store the  // height of the nodes  static int[] height = new int[maxN];   // Adjacency List to store  // the tree  static ArrayList<ArrayList<Integer>> tree = new ArrayList<ArrayList<Integer>>();   // variable to store diameter  // of the tree  static int diameter = 0;   // Function to add edge between  // node u to node v  static void addEdge(int u, int v)  {     // add edge from u to v    tree.get(u).add(v);     // add edge from v to u    tree.get(v).add(u);  }   static 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 (int u : tree.get(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] = Math.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 = Math.max(diameter, height[cur]);    diameter = Math.max(diameter, max1 + max2 + 1);  }   public static void main (String[] args)  {     for(int i = 0; i < maxN; i++)    {      tree.add(new ArrayList<Integer>());           }     // 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);     System.out.println("Diameter of tree is : " +(diameter - 1));  }} // This code is contributed by ab2127.
# C++ implementation to find# diameter of a tree using# DFS in ONE TRAVERSALmaxN = 10005 # The array to store the# height of the nodesheight = [0 for i in range(maxN)] # Adjacency List to store# the treetree = [[] for i in range(maxN)] # variable to store diameter# of the treediameter = 0 # Function to add edge between# node u to node vdef addEdge(u, v):       # add edge from u to v    tree[u].append(v)         # add edge from v to u    tree[v].append(u) def dfs(cur, par):    global diameter         # Variables to store the height of children    # of cur node with maximum heights    max1 = 0    max2 = 0         # going in the adjacency list of the current node    for u in 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]        elif(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# n is the number of nodes in treen = 7 # Adding edges to the treeaddEdge(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 treedfs(1, 0) print("Diameter of tree is :", diameter - 1) # This code is contributed by avanitrachhadiya2155
<script> // Javascript implementation to find// diameter of a tree using// DFS in ONE TRAVERSAL let maxN = 10005; // The array to store the// height of the nodeslet height=new Array(maxN); // Adjacency List to store// the treelet tree=new Array(maxN);for(let i=0;i<maxN;i++){    height[i]=0;    tree[i]=[];} // variable to store diameter// of the treelet diameter = 0; // Function to add edge between// node u to node vfunction addEdge(u,v){    // add edge from u to v    tree[u].push(v);      // add edge from v to u    tree[v].push(u);} function dfs(cur,par){    // Variables to store the height of children    // of cur node with maximum heights    let max1 = 0;    let max2 = 0;      // going in the adjacency list    // of the current node    for (let u=0;u<tree[cur].length;u++) {                  // if that node equals parent discard it        if (tree[cur][u] == par)            continue;          // calling dfs for child node        dfs(tree[cur][u], cur);          // calculating height of nodes        height[cur] = Math.max(height[cur],        height[tree[cur][u]]);          // getting the height of children        // of cur node with maximum height        if (height[tree[cur][u]] >= max1) {            max2 = max1;            max1 = height[tree[cur][u]];        }        else if (height[tree[cur][u]] > max2) {            max2 = height[tree[cur][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 = Math.max(diameter, height[cur]);    diameter = Math.max(diameter, max1 + max2 + 1);} // Driver Code // n is the number of nodes in treelet n = 7; // Adding edges to the treeaddEdge(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 treedfs(1, 0); document.write("Diameter of tree is : " +(diameter - 1))      // This code is contributed by unknown2108 </script>
Diameter of tree is : 4
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