C++ programming Coding DFS and BFS Graph Algorithms

C++ Algorithm – Check if a given graph is tree or not

C++ Algorithm - Check if a given graph is tree or not - Graph Algorithm - Write a function that returns true if a given undirected graph is tree

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++  Programming:

// A C++ Program to check whether a graph is tree or not
#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);
    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;


Graph is Tree
Graph is not Tree
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About the author

Venkatesan Prabu

Venkatesan Prabu

Wikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.

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