Find farthest node from each node in Tree - GeeksforGeeks

09 Aug, 2021
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: 1
Input:  
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++
Java
Python3
C#
Javascript
// 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 edgesvector<int> adj[N]; int lvl[N], dist1[N], dist2[N]; // Add edge between// U and V in treevoid 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 diametervoid 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 nodesvoid 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 diametervoid 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 nodevoid 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 diametervoid 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 codeint main(){    // Function Call    findNodes();     return 0;}
// Java implementation to find the// farthest node from each vertex// of the treeimport 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 treestatic 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 diameterstatic 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 nodesstatic 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 diameterstatic 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 nodestatic 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 diameterstatic 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 Codepublic 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 implementation to find the# farthest node from each vertex# of the tree  # Add edge between# U and V in treedef 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 diameterdef 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 nodesdef 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 diameterdef 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 nodedef 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 diameterdef 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 codeif __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# implementation to find the// farthest node from each vertex// of the treeusing System;using System.Collections.Generic; class GFG{ static int N = 10000; // Adjacency list to store edgesstatic 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 treestatic 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 diameterstatic 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 nodesstatic 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 diameterstatic 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 nodestatic 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 diameterstatic 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 Codepublic 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
<script>// Javascript implementation to find the// farthest node from each vertex// of the treelet N = 10000; // Adjacency list to store edgeslet 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 treefunction 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 diameterfunction 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 nodesfunction 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 diameterfunction 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 nodefunction 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 diameterfunction 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 Codefor(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).
sanjeev2552
GauravRajput1
mohit kumar 29
patel2127
pankajsharmagfg
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Algorithms
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