# C++ Program to Check if an UnDirected Graph is a Tree or Not Using DFS

C++Server Side ProgrammingProgramming

A graph is a tree if it does not contain any cycle. This is a C++ program to check whether an undirected graph is tree or not.

## Algorithm

Begin
function cyclicUtil() :
A) Mark the current node as visited.
B) Recur for all the vertices adjacent to this vertex.
C) If an adjacent is not visited, then recur for that adjacent.
D) If an adjacent is visited and not parent of current vertex, then there is a cycle.
End
Begin
function cyclic():
A) Mark all the vertices as not visited and not part of recursion stack.
B) Call the recursive function cyclicUtil() to detect cycle in differentDFS trees.
End

## Example

#include<iostream>
#include <list>
#include <limits.h>
using namespace std;
class G {
int n;
bool CyclicUtil(int v, bool visited[], int par);
public:
G(int n); //constructor
bool cyclic();
};
G::G(int n) {
this->n = n;
}
}
//recusive function uses visited[] to detect cycle in subgraph reachable from vertex v:
bool G::CyclicUtil(int v, bool visited[], int par) {
visited[v] = true; // Mark the current node as visited
// Recur for all the vertices adjacent to this vertex
list<int>::iterator i;
if (!visited[*i]) //If an adjacent is not visited, then recur for that adjacent {
if (CyclicUtil(*i, visited, v))
return true;
}
// If an adjacent is visited and not parent of current vertex, then there is a cycle.
else if (*i != par)
return true;
}
return false;
}
// to check if the graph is tree or not.
bool G ::cyclic() {
bool *visited = new bool[n]; // Mark all the vertices as not visited and not part of recursion stack
for (int i = 0; i < n; i++)
visited[i] = false;
// Call the recursive function CyclicUtil() to detect cycle in different DFS trees
for (int u = 0; u < n; u++)
if (!visited[u])
if (CyclicUtil(u, visited, -1))
return true;
return false;
}
int main() {
G g1(4);
}
Undirected Graph is a tree