Diameter of an N-ary tree
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 <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;} |
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Java
// 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 |
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Python3
# 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 |
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C#
// 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 |
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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 <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];// varaiable 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;} |
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Output
Diameter of tree is : 4
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