Find farthest node from each node in Tree
Given a Tree, the task is to find the farthest node from each node to another node in this tree.
Example:
Input: Given Adjacency List of Below Tree:
Output:
Farthest node from node 1: 6
Farthest node from node 2: 6
Farthest node from node 3: 6
Farthest node from node 4: 6
Farthest node from node 5: 1
Farthest node from node 6: 1Input:
Output:
Farthest node from node 1: 4
Farthest node from node 2: 7
Farthest node from node 3: 4
Farthest node from node 4: 7
Farthest node from node 5: 7
Farthest node from node 6: 4
Farthest node from node 7: 4
Approach:
First, we have to find two end vertices of the diameter and to find that, we will choose an arbitrary vertex and find the farthest node from this arbitrary vertex and this node will be one end of the diameter and then make it root to find farthest node from it, which will be the other end of diameter. Now for each node, the farthest node will be one of these two end vertices of the diameter of the tree.
Why it works?
Let x and y are the two end vertices of the diameter of the tree and a random vertex is u. Let the farthest vertex from u is v, not x or y. As v is the farthest from u then a new diameter will form having end vertices as x, v or y, v which has greater length but a tree has a unique length of the diameter, so it is not possible and the farthest vertex from u must be x or y.
Below is the implementation of above approach:
C++
// C++ implementation to find the // farthest node from each vertex // of the tree #include <bits/stdc++.h> using namespace std; #define N 10000 // Adjacency list to store edges vector< int > adj[N]; int lvl[N], dist1[N], dist2[N]; // Add edge between // U and V in tree void AddEdge( int u, int v) { // Edge from U to V adj[u].push_back(v); // Edge from V to U adj[v].push_back(u); } int end1, end2, maxi; // DFS to find the first // End Node of diameter void findFirstEnd( int u, int p) { // Calculating level of nodes lvl[u] = 1 + lvl[p]; if (lvl[u] > maxi) { maxi = lvl[u]; end1 = u; } for ( int i = 0; i < adj[u].size(); i++) { // Go in opposite // direction of parent if (adj[u][i] != p) { findFirstEnd(adj[u][i], u); } } } // Function to clear the levels // of the nodes void clear( int n) { // set all value of lvl[] // to 0 for next dfs for ( int i = 0; i <= n; i++) { lvl[i] = 0; } // Set maximum with 0 maxi = 0; dist1[0] = dist2[0] = -1; } // DFS will calculate second // end of the diameter void findSecondEnd( int u, int p) { // Calculating level of nodes lvl[u] = 1 + lvl[p]; if (lvl[u] > maxi) { maxi = lvl[u]; // Store the node with // maximum depth from end1 end2 = u; } for ( int i = 0; i < adj[u].size(); i++) { // Go in opposite // direction of parent if (adj[u][i] != p) { findSecondEnd(adj[u][i], u); } } } // Function to find the distance // of the farthest distant node void findDistancefromFirst( int u, int p) { // Storing distance from // end1 to node u dist1[u] = 1 + dist1[p]; for ( int i = 0; i < adj[u].size(); i++) { if (adj[u][i] != p) { findDistancefromFirst(adj[u][i], u); } } } // Function to find the distance // of nodes from second end of diameter void findDistancefromSecond( int u, int p) { // storing distance from end2 to node u dist2[u] = 1 + dist2[p]; for ( int i = 0; i < adj[u].size(); i++) { if (adj[u][i] != p) { findDistancefromSecond(adj[u][i], u); } } } void findNodes(){ int n = 5; // Joining Edge between two // nodes of the tree AddEdge(1, 2); AddEdge(1, 3); AddEdge(3, 4); AddEdge(3, 5); // Find the one end of // the diameter of tree findFirstEnd(1, 0); clear(n); // Find the other end // of the diameter of tree findSecondEnd(end1, 0); // Find the distance // to each node from end1 findDistancefromFirst(end1, 0); // Find the distance to // each node from end2 findDistancefromSecond(end2, 0); for ( int i = 1; i <= n; i++) { int x = dist1[i]; int y = dist2[i]; // Comparing distance between // the two ends of diameter if (x >= y) { cout << end1 << ' ' ; } else { cout << end2 << ' ' ; } } } // Driver code int main() { // Function Call findNodes(); return 0; } |
Java
// Java implementation to find the // farthest node from each vertex // of the tree import java.util.*; class GFG{ static int N = 10000 ; // Adjacency list to store edges @SuppressWarnings ( "unchecked" ) static Vector<Integer>[] adj = new Vector[N]; static int [] lvl = new int [N], dist1 = new int [N], dist2 = new int [N]; // Add edge between // U and V in tree static void AddEdge( int u, int v) { // Edge from U to V adj[u].add(v); // Edge from V to U adj[v].add(u); } static int end1, end2, maxi; // DFS to find the first // End Node of diameter static void findFirstEnd( int u, int p) { // Calculating level of nodes lvl[u] = 1 + lvl[p]; if (lvl[u] > maxi) { maxi = lvl[u]; end1 = u; } for ( int i = 0 ; i < adj[u].size(); i++) { // Go in opposite // direction of parent if (adj[u].elementAt(i) != p) { findFirstEnd(adj[u].elementAt(i), u); } } } // Function to clear the levels // of the nodes static void clear( int n) { // Set all value of lvl[] // to 0 for next dfs for ( int i = 0 ; i <= n; i++) { lvl[i] = 0 ; } // Set maximum with 0 maxi = 0 ; dist1[ 0 ] = dist2[ 0 ] = - 1 ; } // DFS will calculate second // end of the diameter static void findSecondEnd( int u, int p) { // Calculating level of nodes lvl[u] = 1 + lvl[p]; if (lvl[u] > maxi) { maxi = lvl[u]; // Store the node with // maximum depth from end1 end2 = u; } for ( int i = 0 ; i < adj[u].size(); i++) { // Go in opposite // direction of parent if (adj[u].elementAt(i) != p) { findSecondEnd(adj[u].elementAt(i), u); } } } // Function to find the distance // of the farthest distant node static void findDistancefromFirst( int u, int p) { // Storing distance from // end1 to node u dist1[u] = 1 + dist1[p]; for ( int i = 0 ; i < adj[u].size(); i++) { if (adj[u].elementAt(i) != p) { findDistancefromFirst( adj[u].elementAt(i), u); } } } // Function to find the distance // of nodes from second end of diameter static void findDistancefromSecond( int u, int p) { // Storing distance from end2 to node u dist2[u] = 1 + dist2[p]; for ( int i = 0 ; i < adj[u].size(); i++) { if (adj[u].elementAt(i) != p) { findDistancefromSecond( adj[u].elementAt(i), u); } } } static void findNodes() { int n = 5 ; // Joining Edge between two // nodes of the tree AddEdge( 1 , 2 ); AddEdge( 1 , 3 ); AddEdge( 3 , 4 ); AddEdge( 3 , 5 ); // Find the one end of // the diameter of tree findFirstEnd( 1 , 0 ); clear(n); // Find the other end // of the diameter of tree findSecondEnd(end1, 0 ); // Find the distance // to each node from end1 findDistancefromFirst(end1, 0 ); // Find the distance to // each node from end2 findDistancefromSecond(end2, 0 ); for ( int i = 1 ; i <= n; i++) { int x = dist1[i]; int y = dist2[i]; // Comparing distance between // the two ends of diameter if (x >= y) { System.out.print(end1 + " " ); } else { System.out.print(end2 + " " ); } } } // Driver Code public static void main(String[] args) { for ( int i = 0 ; i < N; i++) { adj[i] = new Vector<>(); } // Function call findNodes(); } } // This code is contributed by sanjeev2552 |
Python3
# Python3 implementation to find the # farthest node from each vertex # of the tree # Add edge between # U and V in tree def AddEdge(u, v): global adj # Edge from U to V adj[u].append(v) # Edge from V to U adj[v].append(u) # DFS to find the first # End Node of diameter def findFirstEnd(u, p): global lvl, adj, end1, maxi # Calculating level of nodes lvl[u] = 1 + lvl[p] if (lvl[u] > maxi): maxi = lvl[u] end1 = u for i in range ( len (adj[u])): # Go in opposite # direction of parent if (adj[u][i] ! = p): findFirstEnd(adj[u][i], u) # Function to clear the levels # of the nodes def clear(n): global lvl, dist1, dist2 # Set all value of lvl[] # to 0 for next dfs for i in range (n + 1 ): lvl[i] = 0 # Set maximum with 0 maxi = 0 dist1[ 0 ] = dist2[ 0 ] = - 1 # DFS will calculate second # end of the diameter def findSecondEnd(u, p): global lvl, adj, maxi, end2 # Calculating level of nodes lvl[u] = 1 + lvl[p] if (lvl[u] > maxi): maxi = lvl[u] # Store the node with # maximum depth from end1 end2 = u for i in range ( len (adj[u])): # Go in opposite # direction of parent if (adj[u][i] ! = p): findSecondEnd(adj[u][i], u) # Function to find the distance # of the farthest distant node def findDistancefromFirst(u, p): global dist1, adj # Storing distance from # end1 to node u dist1[u] = 1 + dist1[p] for i in range ( len (adj[u])): if (adj[u][i] ! = p): findDistancefromFirst(adj[u][i], u) # Function to find the distance # of nodes from second end of diameter def findDistancefromSecond(u, p): global dist2, adj # Storing distance from end2 to node u dist2[u] = 1 + dist2[p] for i in range ( len (adj[u])): if (adj[u][i] ! = p): findDistancefromSecond(adj[u][i], u) def findNodes(): global adj, lvl, dist1, dist2, end1, end2, maxi n = 5 # Joining Edge between two # nodes of the tree AddEdge( 1 , 2 ) AddEdge( 1 , 3 ) AddEdge( 3 , 4 ) AddEdge( 3 , 5 ) # Find the one end of # the diameter of tree findFirstEnd( 1 , 0 ) clear(n) # Find the other end # of the diameter of tree findSecondEnd(end1, 0 ) # Find the distance # to each node from end1 findDistancefromFirst(end1, 0 ) # Find the distance to # each node from end2 findDistancefromSecond(end2, 0 ) for i in range ( 1 , n + 1 ): x = dist1[i] y = dist2[i] # Comparing distance between # the two ends of diameter if (x > = y): print (end1, end = " " ) else : print (end2, end = " " ) # Driver code if __name__ = = '__main__' : adj = [[] for i in range ( 10000 )] lvl = [ 0 for i in range ( 10000 )] dist1 = [ - 1 for i in range ( 10000 )] dist2 = [ - 1 for i in range ( 10000 )] end1, end2, maxi = 0 , 0 , 0 # Function Call findNodes() # This code is contributed by mohit kumar 29 |
C#
// C# implementation to find the // farthest node from each vertex // of the tree using System; using System.Collections.Generic; class GFG{ static int N = 10000; // Adjacency list to store edges static List< int >[] adj = new List< int >[N]; static int [] lvl = new int [N], dist1 = new int [N], dist2 = new int [N]; // Add edge between // U and V in tree static void AddEdge( int u, int v) { // Edge from U to V adj[u].Add(v); // Edge from V to U adj[v].Add(u); } static int end1, end2, maxi; // DFS to find the first // End Node of diameter static void findFirstEnd( int u, int p) { // Calculating level of nodes lvl[u] = 1 + lvl[p]; if (lvl[u] > maxi) { maxi = lvl[u]; end1 = u; } for ( int i = 0; i < adj[u].Count; i++) { // Go in opposite // direction of parent if (adj[u][i] != p) { findFirstEnd(adj[u][i], u); } } } // Function to clear the levels // of the nodes static void clear( int n) { // Set all value of lvl[] // to 0 for next dfs for ( int i = 0; i <= n; i++) { lvl[i] = 0; } // Set maximum with 0 maxi = 0; dist1[0] = dist2[0] = -1; } // DFS will calculate second // end of the diameter static void findSecondEnd( int u, int p) { // Calculating level of nodes lvl[u] = 1 + lvl[p]; if (lvl[u] > maxi) { maxi = lvl[u]; // Store the node with // maximum depth from end1 end2 = u; } for ( int i = 0; i < adj[u].Count; i++) { // Go in opposite // direction of parent if (adj[u][i] != p) { findSecondEnd(adj[u][i], u); } } } // Function to find the distance // of the farthest distant node static void findDistancefromFirst( int u, int p) { // Storing distance from // end1 to node u dist1[u] = 1 + dist1[p]; for ( int i = 0; i < adj[u].Count; i++) { if (adj[u][i] != p) { findDistancefromFirst(adj[u][i], u); } } } // Function to find the distance // of nodes from second end of diameter static void findDistancefromSecond( int u, int p) { // Storing distance from end2 to node u dist2[u] = 1 + dist2[p]; for ( int i = 0; i < adj[u].Count; i++) { if (adj[u][i] != p) { findDistancefromSecond(adj[u][i], u); } } } static void findNodes() { int n = 5; // Joining Edge between two // nodes of the tree AddEdge(1, 2); AddEdge(1, 3); AddEdge(3, 4); AddEdge(3, 5); // Find the one end of // the diameter of tree findFirstEnd(1, 0); clear(n); // Find the other end // of the diameter of tree findSecondEnd(end1, 0); // Find the distance // to each node from end1 findDistancefromFirst(end1, 0); // Find the distance to // each node from end2 findDistancefromSecond(end2, 0); for ( int i = 1; i <= n; i++) { int x = dist1[i]; int y = dist2[i]; // Comparing distance between // the two ends of diameter if (x >= y) { Console.Write(end1 + " " ); } else { Console.Write(end2 + " " ); } } } // Driver Code public static void Main(String[] args) { for ( int i = 0; i < N; i++) { adj[i] = new List< int >(); } // Function call findNodes(); } } // This code is contributed by gauravrajput1 |
Javascript
<script> // Javascript implementation to find the // farthest node from each vertex // of the tree let N = 10000; // Adjacency list to store edges let adj = new Array(N); let lvl = new Array(N); let dist1 = new Array(N); let dist2 = new Array(N); let end = 0, end2 = 0, maxi = 0; for (let i = 0; i < N; i++) { lvl[i] = 0; dist1[i] = -1; dist2[i] = -1; } // Add edge between // U and V in tree function AddEdge(u,v) { // Edge from U to V adj[u].push(v); // Edge from V to U adj[v].push(u); } // DFS to find the first // End Node of diameter function findFirstEnd(u,p) { // Calculating level of nodes lvl[u] = 1 + lvl[p]; if (lvl[u] > maxi) { maxi = lvl[u]; end1 = u; } for (let i = 0; i < adj[u].length; i++) { // Go in opposite // direction of parent if (adj[u][i] != p) { findFirstEnd(adj[u][i], u); } } } // Function to clear the levels // of the nodes function clear(n) { // Set all value of lvl[] // to 0 for next dfs for (let i = 0; i <= n; i++) { lvl[i] = 0; } // Set maximum with 0 maxi = 0; dist1[0] = dist2[0] = -1; } // DFS will calculate second // end of the diameter function findSecondEnd(u,p) { // Calculating level of nodes lvl[u] = 1 + lvl[p]; if (lvl[u] > maxi) { maxi = lvl[u]; // Store the node with // maximum depth from end1 end2 = u; } for (let i = 0; i < adj[u].length; i++) { // Go in opposite // direction of parent if (adj[u][i] != p) { findSecondEnd(adj[u][i], u); } } } // Function to find the distance // of the farthest distant node function findDistancefromFirst(u,p) { // Storing distance from // end1 to node u dist1[u] = 1 + dist1[p]; for (let i = 0; i < adj[u].length; i++) { if (adj[u][i] != p) { findDistancefromFirst( adj[u][i], u); } } } // Function to find the distance // of nodes from second end of diameter function findDistancefromSecond(u,p) { // Storing distance from end2 to node u dist2[u] = 1 + dist2[p]; for (let i = 0; i < adj[u].length; i++) { if (adj[u][i] != p) { findDistancefromSecond( adj[u][i], u); } } } function findNodes() { let n = 5; // Joining Edge between two // nodes of the tree AddEdge(1, 2); AddEdge(1, 3); AddEdge(3, 4); AddEdge(3, 5); // Find the one end of // the diameter of tree findFirstEnd(1, 0); clear(n); // Find the other end // of the diameter of tree findSecondEnd(end1, 0); // Find the distance // to each node from end1 findDistancefromFirst(end1, 0); // Find the distance to // each node from end2 findDistancefromSecond(end2, 0); for (let i = 1; i <= n; i++) { let x = dist1[i]; let y = dist2[i]; // Comparing distance between // the two ends of diameter if (x >= y) { document.write(end1 + " " ); } else { document.write(end2 + " " ); } } } // Driver Code for (let i = 0; i < N; i++) { adj[i] = []; } // Function call findNodes(); // This code is contributed by patel2127 </script> |
4 4 2 2 2
Time Complexity: O(V+E), where V is the number of vertices and E is the number of edges.
Auxiliary Space: O(V + E).