Prerequisites:
Given a N-ary tree with N nodes numbered from 0 to N-1 and a list of undirected edges, the task is to find the node(s) at the center of the given tree.
Eccentricity: The eccentricity of any vertex V in a given tree is the maximum distance between the given vertex V and any other vertex of the tree.
Center: The center of a tree is the vertex having the minimum eccentricity. Hence, it means that in order to find the center we have to minimize this eccentricity.
Examples:
Input: N = 4, Edges[] = { (1, 0), (1, 2), (1, 3)}
Output: 1
Explanation:
Input: N = 6, Edges[] = { (0, 3), (1, 3), (2, 3), (4, 3), (5, 4)}
Output: 3, 4
Explanation:
Approach: It can be observed that the path of maximum eccentricity is the diameter of the tree. Hence, the center of the tree diameter will be the center of the tree as well.
Proof:
- For example, Let’s consider a case where the longest path consists of odd number of vertices. Let the longest path be X —— O ——– Y where X and Y are the two endpoints of the path and O is the middle vertex.
- For a contradiction, if the center of the tree is not O but some other vertex O’, then at least one of the following two statements must be true.
- Path XO’ is strictly longer than path XO
- Path YO’ is strictly longer than path YO
- This means O’ will not satisfy the condition of minimum eccentricity. Hence by contradiction, we have proved that the center of the tree is actually the center of the diameter path.
- Now if the diameter consists odd number of nodes, then there exists only 1 center (also known as Central Tree).
- If diameter consists of even number of nodes, then there are 2 center nodes(also known as Bi-central Tree).
Below is the implementation of the above approach:
C++
// C++ implementation of// the above approach#include <bits/stdc++.h>using namespace std;// To create treemap<int, vector<int> > tree;// Function to store the path// from given vertex to the target// vertex in a vector pathbool getDiameterPath(int vertex, int targetVertex, int parent, vector<int>& path){ // If the target node is found, // push it into path vector if (vertex == targetVertex) { path.push_back(vertex); return true; } for (auto i : tree[vertex]) { // To prevent visiting a // node already visited if (i == parent) continue; // Recursive call to the neighbours // of current node inorder // to get the path if (getDiameterPath(i, targetVertex, vertex, path)) { path.push_back(vertex); return true; } } return false;}// Function to obtain and return the// farthest node from a given vertexvoid farthestNode(int vertex, int parent, int height, int& maxHeight, int& maxHeightNode){ // If the current height is maximum // so far, then save the current node if (height > maxHeight) { maxHeight = height; maxHeightNode = vertex; } // Iterate over all the neighbours // of current node for (auto i : tree[vertex]) { // This is to prevent visiting // a already visited node if (i == parent) continue; // Next call will be at 1 height // higher than our current height farthestNode(i, vertex, height + 1, maxHeight, maxHeightNode); }}// Function to add edgesvoid addedge(int a, int b){ tree[a].push_back(b); tree[b].push_back(a);}void FindCenter(int n){ // Now we will find the 1st farthest // node from 0(any arbitrary node) // Perform DFS from 0 and update // the maxHeightNode to obtain // the farthest node from 0 // Reset to -1 int maxHeight = -1; // Reset to -1 int maxHeightNode = -1; farthestNode(0, -1, 0, maxHeight, maxHeightNode); // Stores one end of the diameter int leaf1 = maxHeightNode; // Similarly the other end of // the diameter // Reset the maxHeight maxHeight = -1; farthestNode(maxHeightNode, -1, 0, maxHeight, maxHeightNode); // Stores the second end // of the diameter int leaf2 = maxHeightNode; // Store the diameter into // the vector path vector<int> path; // Diameter is equal to the // path between the two farthest // nodes leaf1 and leaf2 getDiameterPath(leaf1, leaf2, -1, path); int pathSize = path.size(); if (pathSize % 2) { cout << path[pathSize / 2] << endl; } else { cout << path[pathSize / 2] << ", " << path[(pathSize - 1) / 2] << endl; }}// Driver Codeint main(){ int N = 4; addedge(1, 0); addedge(1, 2); addedge(1, 3); FindCenter(N); return 0;} |
Java
// Java implementation of // the above approach import java.util.*;class GFG{// To create treestatic Map<Integer, ArrayList<Integer>> tree;static ArrayList<Integer> path;static int maxHeight, maxHeightNode;// Function to store the path// from given vertex to the target// vertex in a vector pathstatic boolean getDiameterPath(int vertex, int targetVertex, int parent, ArrayList<Integer> path){ // If the target node is found, // push it into path vector if (vertex == targetVertex) { path.add(vertex); return true; } for(Integer i : tree.get(vertex)) { // To prevent visiting a // node already visited if (i == parent) continue; // Recursive call to the neighbours // of current node inorder // to get the path if (getDiameterPath(i, targetVertex, vertex, path)) { path.add(vertex); return true; } } return false;}// Function to obtain and return the// farthest node from a given vertexstatic void farthestNode(int vertex, int parent, int height){ // If the current height is maximum // so far, then save the current node if (height > maxHeight) { maxHeight = height; maxHeightNode = vertex; } // Iterate over all the neighbours // of current node if (tree.get(vertex) != null) for(Integer i : tree.get(vertex)) { // This is to prevent visiting // a already visited node if (i == parent) continue; // Next call will be at 1 height // higher than our current height farthestNode(i, vertex, height + 1); }}// Function to add edgesstatic void addedge(int a, int b){ if (tree.get(a) == null) tree.put(a, new ArrayList<>()); tree.get(a).add(b); if (tree.get(b) == null) tree.put(b, new ArrayList<>()); tree.get(b).add(a);}static void FindCenter(int n){ // Now we will find the 1st farthest // node from 0(any arbitrary node) // Perform DFS from 0 and update // the maxHeightNode to obtain // the farthest node from 0 // Reset to -1 maxHeight = -1; // Reset to -1 maxHeightNode = -1; farthestNode(0, -1, 0); // Stores one end of the diameter int leaf1 = maxHeightNode; // Similarly the other end of // the diameter // Reset the maxHeight maxHeight = -1; farthestNode(maxHeightNode, -1, 0); // Stores the second end // of the diameter int leaf2 = maxHeightNode; // Store the diameter into // the vector path path = new ArrayList<>(); // Diameter is equal to the // path between the two farthest // nodes leaf1 and leaf2 getDiameterPath(leaf1, leaf2, -1, path); int pathSize = path.size(); if (pathSize % 2 == 1) { System.out.println(path.get(pathSize / 2)); } else { System.out.println(path.get(pathSize / 2) + ", " + path.get((pathSize - 1) / 2)); }}// Driver codepublic static void main(String[] args){ int N = 4; tree = new HashMap<>(); addedge(1, 0); addedge(1, 2); addedge(1, 3); FindCenter(N);}}// This code is contributed by offbeat |
Python3
# Python3 implementation of the above approach# To create treetree = {}path = []maxHeight, maxHeightNode = -1, -1# Function to store the path# from given vertex to the target# vertex in a vector pathdef getDiameterPath(vertex, targetVertex, parent, path): # If the target node is found, # push it into path vector if (vertex == targetVertex): path.append(vertex) return True for i in range(len(tree[vertex])): # To prevent visiting a # node already visited if (tree[vertex][i] == parent): continue # Recursive call to the neighbours # of current node inorder # to get the path if (getDiameterPath(tree[vertex][i], targetVertex, vertex, path)): path.append(vertex) return True return False# Function to obtain and return the# farthest node from a given vertexdef farthestNode(vertex, parent, height): global maxHeight, maxHeightNode # If the current height is maximum # so far, then save the current node if (height > maxHeight): maxHeight = height maxHeightNode = vertex # Iterate over all the neighbours # of current node if (vertex in tree): for i in range(len(tree[vertex])): # This is to prevent visiting # a already visited node if (tree[vertex][i] == parent): continue # Next call will be at 1 height # higher than our current height farthestNode(tree[vertex][i], vertex, height + 1)# Function to add edgesdef addedge(a, b): if (a not in tree): tree[a] = [] tree[a].append(b) if (b not in tree): tree[b] = [] tree[b].append(a)def FindCenter(n): # Now we will find the 1st farthest # node from 0(any arbitrary node) # Perform DFS from 0 and update # the maxHeightNode to obtain # the farthest node from 0 # Reset to -1 maxHeight = -1 # Reset to -1 maxHeightNode = -1 farthestNode(0, -1, 0) # Stores one end of the diameter leaf1 = maxHeightNode # Similarly the other end of # the diameter # Reset the maxHeight maxHeight = -1 farthestNode(maxHeightNode, -1, 0) # Stores the second end # of the diameter leaf2 = maxHeightNode # Store the diameter into # the vector path path = [] # Diameter is equal to the # path between the two farthest # nodes leaf1 and leaf2 getDiameterPath(leaf1, leaf2, -1, path) pathSize = len(path) if (pathSize % 2 == 1): print(path[int(pathSize / 2)]*-1) else: print(path[int(pathSize / 2)], ", ", path[int((pathSize - 1) / 2)], sep = "", end = "")N = 4 tree = {}addedge(1, 0)addedge(1, 2)addedge(1, 3)FindCenter(N)# This code is contributed by suresh07. |
C#
// C# implementation of// the above approachusing System;using System.Collections.Generic;class GFG { // To create tree static Dictionary<int, List<int>> tree; static List<int> path; static int maxHeight, maxHeightNode; // Function to store the path // from given vertex to the target // vertex in a vector path static bool getDiameterPath(int vertex, int targetVertex, int parent, List<int> path) { // If the target node is found, // push it into path vector if (vertex == targetVertex) { path.Add(vertex); return true; } foreach(int i in tree[vertex]) { // To prevent visiting a // node already visited if (i == parent) continue; // Recursive call to the neighbours // of current node inorder // to get the path if (getDiameterPath(i, targetVertex, vertex, path)) { path.Add(vertex); return true; } } return false; } // Function to obtain and return the // farthest node from a given vertex static void farthestNode(int vertex, int parent, int height) { // If the current height is maximum // so far, then save the current node if (height > maxHeight) { maxHeight = height; maxHeightNode = vertex; } // Iterate over all the neighbours // of current node if (tree.ContainsKey(vertex) && tree[vertex].Count > 0) { foreach(int i in tree[vertex]) { // This is to prevent visiting // a already visited node if (i == parent) continue; // Next call will be at 1 height // higher than our current height farthestNode(i, vertex, height + 1); } } } // Function to add edges static void addedge(int a, int b) { if (!tree.ContainsKey(a)) tree[a] = new List<int>(); tree[a].Add(b); if (!tree.ContainsKey(b)) tree[b] = new List<int>(); tree[b].Add(a); } static void FindCenter(int n) { // Now we will find the 1st farthest // node from 0(any arbitrary node) // Perform DFS from 0 and update // the maxHeightNode to obtain // the farthest node from 0 // Reset to -1 maxHeight = -1; // Reset to -1 maxHeightNode = -1; farthestNode(0, -1, 0); // Stores one end of the diameter int leaf1 = maxHeightNode; // Similarly the other end of // the diameter // Reset the maxHeight maxHeight = -1; farthestNode(maxHeightNode, -1, 0); // Stores the second end // of the diameter int leaf2 = maxHeightNode; // Store the diameter into // the vector path path = new List<int>(); // Diameter is equal to the // path between the two farthest // nodes leaf1 and leaf2 getDiameterPath(leaf1, leaf2, -1, path); int pathSize = path.Count; if (pathSize % 2 == 1) { Console.WriteLine(path[pathSize / 2]); } else { Console.WriteLine(path[pathSize / 2] + ", " + path[(pathSize - 1) / 2]); } } static void Main() { int N = 4; tree = new Dictionary<int, List<int>>(); addedge(1, 0); addedge(1, 2); addedge(1, 3); FindCenter(N); }}// This code is contributed by divyesh072019. |
Javascript
<script> // Javascript implementation of the above approach // To create tree let tree; let path; let maxHeight, maxHeightNode; // Function to store the path // from given vertex to the target // vertex in a vector path function getDiameterPath(vertex, targetVertex, parent, path) { // If the target node is found, // push it into path vector if (vertex == targetVertex) { path.push(vertex); return true; } for(let i = 0; i < tree.get(vertex).length; i++) { // To prevent visiting a // node already visited if (tree.get(vertex)[i] == parent) continue; // Recursive call to the neighbours // of current node inorder // to get the path if (getDiameterPath(tree.get(vertex)[i], targetVertex, vertex, path)) { path.push(vertex); return true; } } return false; } // Function to obtain and return the // farthest node from a given vertex function farthestNode(vertex, parent, height) { // If the current height is maximum // so far, then save the current node if (height > maxHeight) { maxHeight = height; maxHeightNode = vertex; } // Iterate over all the neighbours // of current node if (tree.get(vertex) != null) for(let i = 0; i < tree.get(vertex).length; i++) { // This is to prevent visiting // a already visited node if (tree.get(vertex)[i] == parent) continue; // Next call will be at 1 height // higher than our current height farthestNode(tree.get(vertex)[i], vertex, height + 1); } } // Function to add edges function addedge(a, b) { if (tree.get(a) == null) tree.set(a, []); tree.get(a).push(b); if (tree.get(b) == null) tree.set(b, []); tree.get(b).push(a); } function FindCenter(n) { // Now we will find the 1st farthest // node from 0(any arbitrary node) // Perform DFS from 0 and update // the maxHeightNode to obtain // the farthest node from 0 // Reset to -1 maxHeight = -1; // Reset to -1 maxHeightNode = -1; farthestNode(0, -1, 0); // Stores one end of the diameter let leaf1 = maxHeightNode; // Similarly the other end of // the diameter // Reset the maxHeight maxHeight = -1; farthestNode(maxHeightNode, -1, 0); // Stores the second end // of the diameter let leaf2 = maxHeightNode; // Store the diameter into // the vector path path = []; // Diameter is equal to the // path between the two farthest // nodes leaf1 and leaf2 getDiameterPath(leaf1, leaf2, -1, path); let pathSize = path.length; if (pathSize % 2 == 1) { document.write(path[parseInt(pathSize / 2, 10)]); } else { document.write(path[parseInt(pathSize / 2, 10)] + ", " + path[parseInt((pathSize - 1) / 2, 10)]); } } let N = 4; tree = new Map(); addedge(1, 0); addedge(1, 2); addedge(1, 3); FindCenter(N);</script> |
Output:
1
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
