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Euler Circuit in a Directed Graph - GeeksforGeeks
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Euler Circuit in a Directed Graph

Last Updated : 23 Jul, 2025
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Eulerian Path is a path in graph that visits every edge exactly once. Eulerian Circuit is an Eulerian Path which starts and ends on the same vertex. 

A graph is said to be eulerian if it has a eulerian cycle. We have discussed eulerian circuit for an undirected graph. In this post, the same is discussed for a directed graph.

For example, the following graph has eulerian cycle as {1, 0, 3, 4, 0, 2, 1} 

SCC


How to check if a directed graph is eulerian? 

A directed graph has an eulerian cycle if following conditions are true

  1. All vertices with nonzero degree belong to a single strongly connected component
  2. In degree is equal to the out degree for every vertex.

We can detect singly connected component using Kosaraju’s DFS based simple algorithm

To compare in degree and out-degree, we need to store in degree and out-degree of every vertex. Out degree can be obtained by the size of an adjacency list. In degree can be stored by creating an array of size equal to the number of vertices. 

Following implementations of above approach. 

C++
// A C++ program to check if a given directed graph is Eulerian or not
#include<iostream>
#include <list>
#define CHARS 26
using namespace std;

// A class that represents an undirected graph
class Graph
{
    int V;    // No. of vertices
    list<int> *adj;    // A dynamic array of adjacency lists
    int *in;
public:
    // Constructor and destructor
    Graph(int V);
    ~Graph()   { delete [] adj; delete [] in; }

    // function to add an edge to graph
    void addEdge(int v, int w) { adj[v].push_back(w);  (in[w])++; }

    // Method to check if this graph is Eulerian or not
    bool isEulerianCycle();

    // Method to check if all non-zero degree vertices are connected
    bool isSC();

    // Function to do DFS starting from v. Used in isConnected();
    void DFSUtil(int v, bool visited[]);

    Graph getTranspose();
};

Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
    in = new int[V];
    for (int i = 0; i < V; i++)
       in[i] = 0;
}

/* This function returns true if the directed graph has a eulerian
   cycle, otherwise returns false  */
bool Graph::isEulerianCycle()
{
    // Check if all non-zero degree vertices are connected
    if (isSC() == false)
        return false;

    // Check if in degree and out degree of every vertex is same
    for (int i = 0; i < V; i++)
        if (adj[i].size() != in[i])
            return false;

    return true;
}

// A recursive function to do DFS starting from v
void Graph::DFSUtil(int v, bool visited[])
{
    // Mark the current node as visited and print it
    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 (!visited[*i])
            DFSUtil(*i, visited);
}

// Function that returns reverse (or transpose) of this graph
// This function is needed in isSC()
Graph Graph::getTranspose()
{
    Graph g(V);
    for (int v = 0; v < V; v++)
    {
        // Recur for all the vertices adjacent to this vertex
        list<int>::iterator i;
        for(i = adj[v].begin(); i != adj[v].end(); ++i)
        {
            g.adj[*i].push_back(v);
            (g.in[v])++;
        }
    }
    return g;
}

// This function returns true if all non-zero degree vertices of 
// graph are strongly connected (Please refer 
// https://www.geeksforgeeks.org/dsa/connectivity-in-a-directed-graph/ )
bool Graph::isSC()
{
    // Mark all the vertices as not visited (For first DFS)
    bool visited[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;

    // Find the first vertex with non-zero degree
    int n;
    for (n = 0; n < V; n++)
        if (adj[n].size() > 0)
          break;

    // Do DFS traversal starting from first non zero degrees vertex.
    DFSUtil(n, visited);

     // If DFS traversal doesn't visit all vertices, then return false.
    for (int i = 0; i < V; i++)
        if (adj[i].size() > 0 && visited[i] == false)
              return false;

    // Create a reversed graph
    Graph gr = getTranspose();

    // Mark all the vertices as not visited (For second DFS)
    for (int i = 0; i < V; i++)
        visited[i] = false;

    // Do DFS for reversed graph starting from first vertex.
    // Starting Vertex must be same starting point of first DFS
    gr.DFSUtil(n, visited);

    // If all vertices are not visited in second DFS, then
    // return false
    for (int i = 0; i < V; i++)
        if (adj[i].size() > 0 && visited[i] == false)
             return false;

    return true;
}

// Driver program to test above functions
int main()
{
    // Create a graph given in the above diagram
    Graph g(5);
    g.addEdge(1, 0);
    g.addEdge(0, 2);
    g.addEdge(2, 1);
    g.addEdge(0, 3);
    g.addEdge(3, 4);
    g.addEdge(4, 0);

    if (g.isEulerianCycle())
       cout << "Given directed graph is eulerian n";
    else
       cout << "Given directed graph is NOT eulerian n";
    return 0;
}
Java Python3 C# JavaScript

Output
Given directed graph is eulerian n

 Time complexity of the above implementation is O(V + E) as Kosaraju’s algorithm takes O(V + E) time. After running Kosaraju’s algorithm we traverse all vertices and compare in degree with out degree which takes O(V) time. 

Auxiliary Space : O(V), since an extra visited array of size V is required.

See following as an application of this. 
Find if the given array of strings can be chained to form a circle.


Euler Circuit in an Undirected Graph
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