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: 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++

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// 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;
}

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Output:

4 4 2 2 2

Time Complexity: O(V+E)

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