# Check if a given graph is tree or not

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

But the following graph is not a tree.

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
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;
}

{
}

// 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;
{
// If an adjacent is not visited, then recur for
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.isTree()? cout << "Graph is Tree\n":
cout << "Graph is not Tree\n";

Graph g2(5);
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

// Constructor
Graph(int v)
{
V = v;
for (int i=0; i<v; ++i)
}

// Function to add an edge into the graph
{
}

// 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
while (it.hasNext())
{
i = it.next();

// If an adjacent is not visited, then recur for
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);
if (g1.isTree())
System.out.println("Graph is Tree");
else
System.out.println("Graph is not Tree");

Graph g2 = new Graph(5);

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)

# 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)
print "Graph is a Tree" if g1.isTree() == True \
else "Graph is a not a Tree"

g2 = Graph(5)
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

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