Check if a given graph is tree or not

3.1

Write a function that returns true if a given undirected graph is tree and false otherwise. For example, the following graph is a tree.

cycleGraph

But the following graph is not a tree.
cycleGraph

An undirected graph is tree if it has following properties.
1) There is no cycle.
2) The graph is connected.

For an undirected graph we can either use BFS or DFS to detect above two properties.

How to detect cycle in an undirected graph?
We can either use BFS or DFS. For every visited vertex ‘v’, if there is an adjacent ‘u’ such that u is already visited and u is not parent of v, then there is a cycle in graph. If we don’t find such an adjacent for any vertex, we say that there is no cycle (See Detect cycle in an undirected graph for more details).

How to check for connectivity?
Since the graph is undirected, we can start BFS or DFS from any vertex and check if all vertices are reachable or not. If all vertices are reachable, then graph is connected, otherwise not.

C++

// A C++ Program to check whether a graph is tree or not
#include<iostream>
#include <list>
#include <limits.h>
using namespace std;

// Class for an undirected graph
class Graph
{
    int V;    // No. of vertices
    list<int> *adj; // Pointer to an array for adjacency lists
    bool isCyclicUtil(int v, bool visited[], int parent);
public:
    Graph(int V);   // Constructor
    void addEdge(int v, int w);   // to add an edge to graph
    bool isTree();   // returns true if graph is tree
};

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); // Add v to w’s list.
}

// A recursive function that uses visited[] and parent to
// detect cycle in subgraph reachable from vertex v.
bool Graph::isCyclicUtil(int v, bool visited[], int parent)
{
    // Mark the current node as visited
    visited[v] = true;

    // Recur for all the vertices adjacent to this vertex
    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
    {
        // If an adjacent is not visited, then recur for 
        // that adjacent
        if (!visited[*i])
        {
           if (isCyclicUtil(*i, visited, v))
              return true;
        }

        // If an adjacent is visited and not parent of current
        // vertex, then there is a cycle.
        else if (*i != parent)
           return true;
    }
    return false;
}

// Returns true if the graph is a tree, else false.
bool Graph::isTree()
{
    // Mark all the vertices as not visited and not part of 
    // recursion stack
    bool *visited = new bool[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;

    // The call to isCyclicUtil serves multiple purposes.
    // It returns true if graph reachable from vertex 0 
    // is cyclcic. It also marks all vertices reachable 
    // from 0.
    if (isCyclicUtil(0, visited, -1))
             return false;

    // If we find a vertex which is not reachable from 0 
    // (not marked by isCyclicUtil(), then we return false
    for (int u = 0; u < V; u++)
        if (!visited[u])
           return false;

    return true;
}

// Driver program to test above functions
int main()
{
    Graph g1(5);
    g1.addEdge(1, 0);
    g1.addEdge(0, 2);
    g1.addEdge(0, 3);
    g1.addEdge(3, 4);
    g1.isTree()? cout << "Graph is Tree\n":
                 cout << "Graph is not Tree\n";

    Graph g2(5);
    g2.addEdge(1, 0);
    g2.addEdge(0, 2);
    g2.addEdge(2, 1);
    g2.addEdge(0, 3);
    g2.addEdge(3, 4);
    g2.isTree()? cout << "Graph is Tree\n":
                 cout << "Graph is not Tree\n";

    return 0;
}

Java

// A Java Program to check whether a graph is tree or not
import java.io.*;
import java.util.*;

// This class represents a directed graph using adjacency
// list representation
class Graph
{
    private int V;   // No. of vertices
    private LinkedList<Integer> adj[]; //Adjacency List

    // Constructor
    Graph(int v)
    {
        V = v;
        adj = new LinkedList[v];
        for (int i=0; i<v; ++i)
            adj[i] = new LinkedList();
    }

    // Function to add an edge into the graph
    void addEdge(int v,int w)
    {
        adj[v].add(w);
        adj[w].add(v);
    }

    // A recursive function that uses visited[] and parent
    // to detect cycle in subgraph reachable from vertex v.
    Boolean isCyclicUtil(int v, Boolean visited[], int parent)
    {
        // Mark the current node as visited
        visited[v] = true;
        Integer i;

        // Recur for all the vertices adjacent to this vertex
        Iterator<Integer> it = adj[v].iterator();
        while (it.hasNext())
        {
            i = it.next();

            // If an adjacent is not visited, then recur for
            // that adjacent
            if (!visited[i])
            {
                if (isCyclicUtil(i, visited, v))
                    return true;
            }

            // If an adjacent is visited and not parent of 
            // current vertex, then there is a cycle.
            else if (i != parent)
               return true;
        }
        return false;
    }

    // Returns true if the graph is a tree, else false.
    Boolean isTree()
    {
        // Mark all the vertices as not visited and not part
        // of recursion stack
        Boolean visited[] = new Boolean[V];
        for (int i = 0; i < V; i++)
            visited[i] = false;

        // The call to isCyclicUtil serves multiple purposes
        // It returns true if graph reachable from vertex 0
        // is cyclcic. It also marks all vertices reachable
        // from 0.
        if (isCyclicUtil(0, visited, -1))
            return false;

        // If we find a vertex which is not reachable from 0
        // (not marked by isCyclicUtil(), then we return false
        for (int u = 0; u < V; u++)
            if (!visited[u])
                return false;

        return true;
    }

    // Driver method
    public static void main(String args[])
    {
        // Create a graph given in the above diagram
        Graph g1 = new Graph(5);
        g1.addEdge(1, 0);
        g1.addEdge(0, 2);
        g1.addEdge(0, 3);
        g1.addEdge(3, 4);
        if (g1.isTree())
            System.out.println("Graph is Tree");
        else
            System.out.println("Graph is not Tree");

        Graph g2 = new Graph(5);
        g2.addEdge(1, 0);
        g2.addEdge(0, 2);
        g2.addEdge(2, 1);
        g2.addEdge(0, 3);
        g2.addEdge(3, 4);

        if (g2.isTree())
            System.out.println("Graph is Tree");
        else
            System.out.println("Graph is not Tree");

    }
}
// This code is contributed by Aakash Hasija

Python

# Python Program to check whether 
# a graph is tree or not

from collections import defaultdict

class Graph():

    def __init__(self, V):
        self.V = V
        self.graph  = defaultdict(list)

    def addEdge(self, v, w):
        # Add w to v ist.
        self.graph[v].append(w) 
        # Add v to w list.
        self.graph[w].append(v) 

    # A recursive function that uses visited[] 
    # and parent to detect cycle in subgraph 
    # reachable from vertex v.
    def isCyclicUtil(self, v, visited, parent):

        # Mark current node as visited
        visited[v] = True

        # Recur for all the vertices adjacent 
        # for this vertex
        for i in self.graph[v]:
            # If an adjacent is not visited, 
            # then recur for that adjacent
            if visited[i] == False:
                if self.isCyclicUtil(i, visited, v) == True:
                    return True

            # If an adjacent is visited and not 
            # parent of current vertex, then there 
            # is a cycle.
            elif i != parent:
                return True

        return False

    # Returns true if the graph is a tree, 
    # else false.
    def isTree(self):
        # Mark all the vertices as not visited 
        # and not part of recursion stack
        visited = [False] * self.V

        # The call to isCyclicUtil serves multiple 
        # purposes. It returns true if graph reachable 
        # from vertex 0 is cyclcic. It also marks 
        # all vertices reachable from 0.
        if self.isCyclicUtil(0, visited, -1) == True:
            return False

        # If we find a vertex which is not reachable
        # from 0 (not marked by isCyclicUtil(), 
        # then we return false
        for i in range(self.V):
            if visited[i] == False:
                return False

        return True

# Driver program to test above functions
g1 = Graph(5)
g1.addEdge(1, 0)
g1.addEdge(0, 2)
g1.addEdge(0, 3)
g1.addEdge(3, 4)
print "Graph is a Tree" if g1.isTree() == True \
                          else "Graph is a not a Tree"

g2 = Graph(5)
g2.addEdge(1, 0)
g2.addEdge(0, 2)
g2.addEdge(2, 1)
g2.addEdge(0, 3)
g2.addEdge(3, 4)
print "Graph is a Tree" if g2.isTree() == True \
                          else "Graph is a not a Tree"
                          
# This code is contributed by Divyanshu Mehta      


Output:
Graph is Tree
Graph is not Tree

Thanks to Vinit Verma for suggesting this problem and initial solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



3.1 Average Difficulty : 3.1/5.0
Based on 19 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.