Longest path in an undirected tree

Given an undirected tree, we need to find the longest path of this tree where a path is defined as a sequence of nodes.
Example:

Input : Below shown Tree using adjacency list 
        representation:
Output : 5
In below tree longest path is of length 5
from node 5 to node 7


This problem is same as diameter of n-ary tree. We have discussed a simple solution here.

In this post, an efficient solution is discussed. We can find longest path using two BFSs. The idea is based on the following fact: If we start BFS from any node x and find a node with the longest distance from x, it must be an end point of the longest path. It can be proved using contradiction. So our algorithm reduces to simple two BFSs. First BFS to find an end point of the longest path and second BFS from this end point to find the actual longest path.
As we can see in above diagram, if we start our BFS from node-0, the node at the farthest distance from it will be node-5, now if we start our BFS from node-5 the node at the farthest distance will be node-7, finally, path from node-5 to node-7 will constitute our longest path.

// C++ program to find longest path of the tree
#include <bits/stdc++.h>
using namespace std;

// This class represents a undirected graph using adjacency list
class Graph
{
    int V;              // No. of vertices
    list<int> *adj;     // Pointer to an array containing
                        // adjacency lists
public:
    Graph(int V);              // Constructor
    void addEdge(int v, int w);// function to add an edge to graph
    void longestPathLength();  // prints longest path of the tree
    pair<int, int> bfs(int u); // function returns maximum distant
                               // node from u with its distance
};

Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}

void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w);    // Add w to v’s list.
    adj[w].push_back(v);    // Since the graph is undirected
}

//  method returns farthest node and its distance from node u
pair<int, int> Graph::bfs(int u)
{
    //  mark all distance with -1
    int dis[V];
    memset(dis, -1, sizeof(dis));

    queue<int> q;
    q.push(u);

    //  distance of u from u will be 0
    dis[u] = 0;

    while (!q.empty())
    {
        int t = q.front();       q.pop();

        //  loop for all adjacent nodes of node-t
        for (auto it = adj[t].begin(); it != adj[t].end(); it++)
        {
            int v = *it;

            // push node into queue only if
            // it is not visited already
            if (dis[v] == -1)
            {
                q.push(v);

                // make distance of v, one more
                // than distance of t
                dis[v] = dis[t] + 1;
            }
        }
    }

    int maxDis = 0;
    int nodeIdx;

    //  get farthest node distance and its index
    for (int i = 0; i < V; i++)
    {
        if (dis[i] > maxDis)
        {
            maxDis = dis[i];
            nodeIdx = i;
        }
    }
    return make_pair(nodeIdx, maxDis);
}

//  method prints longest path of given tree
void Graph::longestPathLength()
{
    pair<int, int> t1, t2;

    // first bfs to find one end point of
    // longest path
    t1 = bfs(0);

    //  second bfs to find actual longest path
    t2 = bfs(t1.first);

    cout << "Longest path is from " << t1.first << " to "
         << t2.first << " of length " << t2.second;
}

// Driver code to test above methods
int main()
{
    // Create a graph given in the example
    Graph g(10);
    g.addEdge(0, 1);
    g.addEdge(1, 2);
    g.addEdge(2, 3);
    g.addEdge(2, 9);
    g.addEdge(2, 4);
    g.addEdge(4, 5);
    g.addEdge(1, 6);
    g.addEdge(6, 7);
    g.addEdge(6, 8);

    g.longestPathLength();
    return 0;
}

Output:

Longest path is from 5 to 7 of length 5

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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