Fleury’s Algorithm for printing Eulerian Path or Circuit

Last Updated : 06 Jun, 2023

Eulerian Path is a path in a graph that visits every edge exactly once. Eulerian Circuit is an Eulerian Path that starts and ends on the same vertex.

We strongly recommend first reading the following post on Euler Path and Circuit. “https://www.geeksforgeeks.org/eulerian-path-and-circuit/”

In the above-mentioned post, we discussed the problem of finding out whether a given graph is Eulerian or not. In this post, an algorithm to print an Eulerian trail or circuit is discussed.

Following is Fleury’s Algorithm for printing the Eulerian trail or cycle

Make sure the graph has either 0 or 2 odd vertices.

If there are 0 odd vertices, start anywhere. If there are 2 odd vertices, start at one of them.

Follow edges one at a time. If you have a choice between a bridge and a non-bridge, always choose the non-bridge.

Stop when you run out of edges.

The idea is, “don’t burn bridges“ so that we can come back to a vertex and traverse the remaining edges. For example, let us consider the following graph.

Euler1

There are two vertices with odd degrees, ‘2’ and ‘3’, and we can start paths from any of them. Let us start the tour from vertex ‘2’.

Euler2

Three edges are going out from vertex ‘2’, which one to pick? We don’t pick the edge ‘2-3’ because that is a bridge (we won’t be able to come back to ‘3’). We can pick any of the remaining two edges. Let us say we pick ‘2-0’. We remove this edge and move to vertex ‘0’.

Eule3

There is only one edge from vertex ‘0’, so we pick it, remove it and move to vertex ‘1’. Euler tour becomes ‘2-0 0-1’.

Euler4

There is only one edge from vertex ‘1’, so we pick it, remove it and move to vertex ‘2’. Euler tour becomes ‘2-0 0-1 1-2’

Euler5

Again there is only one edge from vertex 2, so we pick it, remove it and move to vertex 3. Euler tour becomes ‘2-0 0-1 1-2 2-3’

Euler6

There are no more edges left, so we stop here. Final tour is ‘2-0 0-1 1-2 2-3’.

Following is the C++ implementation of the above algorithm. In the following code, it is assumed that the given graph has an Eulerian trail or Circuit. The main focus is to print an Eulerian trail or circuit. We can use isEulerian() to first check whether there is an Eulerian Trail or Circuit in the given graph.

We first find the starting point which must be an odd vertex (if there are odd vertices) and store it in variable ‘u’. If there are zero odd vertices, we start from vertex ‘0’. We call printEulerUtil() to print Euler tour starting with u. We traverse all adjacent vertices of u, if there is only one adjacent vertex, we immediately consider it. If there are more than one adjacent vertices, we consider an adjacent v only if edge u-v is not a bridge. How to find if a given edge is a bridge? We count several vertices reachable from u. We remove edge u-v and again count the number of reachable vertices from u. If the number of reachable vertices is reduced, then edge u-v is a bridge. To count reachable vertices, we can either use BFS or DFS, we have used DFS in the above code. The function DFSCount(u) returns several vertices reachable from u.

Once an edge is processed (included in the Euler tour), we remove it from the graph. To remove the edge, we replace the vertex entry with -1 in the adjacency list. Note that simply deleting the node may not work as the code is recursive and a parent call may be in the middle of the adjacency list.

C++

// A C++ program print Eulerian Trail in a given Eulerian or
// Semi-Eulerian Graph
#include <algorithm>
#include <iostream>
#include <list>
#include <string.h>
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
public:
    // Constructor and destructor
    Graph(int V)
    {
        this->V = V;
        adj = new list<int>[V];
    }
    ~Graph() { delete[] adj; }
 
    // functions to add and remove edge
    void addEdge(int u, int v)
    {
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    void rmvEdge(int u, int v);
 
    // Methods to print Eulerian tour
    void printEulerTour();
    void printEulerUtil(int s);
 
    // This function returns count of vertices reachable
    // from v. It does DFS
    int DFSCount(int v, bool visited[]);
 
    // Utility function to check if edge u-v is a valid next
    // edge in Eulerian trail or circuit
    bool isValidNextEdge(int u, int v);
};
 
/* The main function that print Eulerian Trail. It first
   finds an odd degree vertex (if there is any) and then
   calls printEulerUtil() to print the path */
void Graph::printEulerTour()
{
    // Find a vertex with odd degree
    int u = 0;
    for (int i = 0; i < V; i++)
        if (adj[i].size() & 1) {
            u = i;
            break;
        }
 
    // Print tour starting from oddv
    printEulerUtil(u);
    cout << endl;
}
 
// Print Euler tour starting from vertex u
void Graph::printEulerUtil(int u)
{
    // Recur for all the vertices adjacent to this vertex
    list<int>::iterator i;
    for (i = adj[u].begin(); i != adj[u].end(); ++i) {
        int v = *i;
 
        // If edge u-v is not removed and it's a a valid
        // next edge
        if (v != -1 && isValidNextEdge(u, v)) {
            cout << u << "-" << v << "  ";
            rmvEdge(u, v);
            printEulerUtil(v);
        }
    }
}
 
// The function to check if edge u-v can be considered as
// next edge in Euler Tout
bool Graph::isValidNextEdge(int u, int v)
{
    // The edge u-v is valid in one of the following two
    // cases:
 
    // 1) If v is the only adjacent vertex of u
    int count = 0; // To store count of adjacent vertices
    list<int>::iterator i;
    for (i = adj[u].begin(); i != adj[u].end(); ++i)
        if (*i != -1)
            count++;
    if (count == 1)
        return true;
 
    // 2) If there are multiple adjacents, then u-v is not a
    // bridge Do following steps to check if u-v is a bridge
 
    // 2.a) count of vertices reachable from u
    bool visited[V];
    memset(visited, false, V);
    int count1 = DFSCount(u, visited);
 
    // 2.b) Remove edge (u, v) and after removing the edge,
    // count vertices reachable from u
    rmvEdge(u, v);
    memset(visited, false, V);
    int count2 = DFSCount(u, visited);
 
    // 2.c) Add the edge back to the graph
    addEdge(u, v);
 
    // 2.d) If count1 is greater, then edge (u, v) is a
    // bridge
    return (count1 > count2) ? false : true;
}
 
// This function removes edge u-v from graph.  It removes
// the edge by replacing adjacent vertex value with -1.
void Graph::rmvEdge(int u, int v)
{
    // Find v in adjacency list of u and replace it with -1
    list<int>::iterator iv
        = find(adj[u].begin(), adj[u].end(), v);
    *iv = -1;
 
    // Find u in adjacency list of v and replace it with -1
    list<int>::iterator iu
        = find(adj[v].begin(), adj[v].end(), u);
    *iu = -1;
}
 
// A DFS based function to count reachable vertices from v
int Graph::DFSCount(int v, bool visited[])
{
    // Mark the current node as visited
    visited[v] = true;
    int count = 1;
 
    // Recur for all vertices adjacent to this vertex
    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
        if (*i != -1 && !visited[*i])
            count += DFSCount(*i, visited);
 
    return count;
}
 
// Driver program to test above function
int main()
{
    // Let us first create and test graphs shown in above
    // figure
    Graph g1(4);
    g1.addEdge(0, 1);
    g1.addEdge(0, 2);
    g1.addEdge(1, 2);
    g1.addEdge(2, 3);
    g1.printEulerTour();
 
    Graph g2(3);
    g2.addEdge(0, 1);
    g2.addEdge(1, 2);
    g2.addEdge(2, 0);
    g2.printEulerTour();
 
    Graph g3(5);
    g3.addEdge(1, 0);
    g3.addEdge(0, 2);
    g3.addEdge(2, 1);
    g3.addEdge(0, 3);
    g3.addEdge(3, 4);
    g3.addEdge(3, 2);
    g3.addEdge(3, 1);
    g3.addEdge(2, 4);
    g3.printEulerTour();
 
    return 0;
}

Java

// A Java program print Eulerian Trail
// in a given Eulerian or Semi-Eulerian Graph
import java.util.ArrayList;
 
// An Undirected graph using
// adjacency list representation
public class Graph {
 
    private int vertices; // No. of vertices
    private ArrayList<Integer>[] adj; // adjacency list
 
    // Constructor
    Graph(int numOfVertices)
    {
        // initialise vertex count
        this.vertices = numOfVertices;
 
        // initialise adjacency list
        initGraph();
    }
 
    // utility method to initialise adjacency list
    @SuppressWarnings("unchecked") private void initGraph()
    {
        adj = new ArrayList[vertices];
        for (int i = 0; i < vertices; i++) {
            adj[i] = new ArrayList<>();
        }
    }
 
    // add edge u-v
    private void addEdge(Integer u, Integer v)
    {
        adj[u].add(v);
        adj[v].add(u);
    }
 
    // This function removes edge u-v from graph.
    private void removeEdge(Integer u, Integer v)
    {
        adj[u].remove(v);
        adj[v].remove(u);
    }
 
    /* The main function that print Eulerian Trail.
       It first finds an odd degree vertex (if there
       is any) and then calls printEulerUtil() to
       print the path */
    private void printEulerTour()
    {
        // Find a vertex with odd degree
        Integer u = 0;
        for (int i = 0; i < vertices; i++) {
            if (adj[i].size() % 2 == 1) {
                u = i;
                break;
            }
        }
 
        // Print tour starting from oddv
        printEulerUtil(u);
        System.out.println();
    }
 
    // Print Euler tour starting from vertex u
    private void printEulerUtil(Integer u)
    {
        // Recur for all the vertices adjacent to this
        // vertex
        for (int i = 0; i < adj[u].size(); i++) {
            Integer v = adj[u].get(i);
            // If edge u-v is a valid next edge
            if (isValidNextEdge(u, v)) {
                System.out.print(u + "-" + v + " ");
 
                // This edge is used so remove it now
                removeEdge(u, v);
                printEulerUtil(v);
            }
        }
    }
 
    // The function to check if edge u-v can be
    // considered as next edge in Euler Tout
    private boolean isValidNextEdge(Integer u, Integer v)
    {
        // The edge u-v is valid in one of the
        // following two cases:
 
        // 1) If v is the only adjacent vertex of u
        // ie size of adjacent vertex list is 1
        if (adj[u].size() == 1) {
            return true;
        }
 
        // 2) If there are multiple adjacents, then
        // u-v is not a bridge Do following steps
        // to check if u-v is a bridge
        // 2.a) count of vertices reachable from u
        boolean[] isVisited = new boolean[this.vertices];
        int count1 = dfsCount(u, isVisited);
 
        // 2.b) Remove edge (u, v) and after removing
        //  the edge, count vertices reachable from u
        removeEdge(u, v);
        isVisited = new boolean[this.vertices];
        int count2 = dfsCount(u, isVisited);
 
        // 2.c) Add the edge back to the graph
        addEdge(u, v);
        return (count1 > count2) ? false : true;
    }
 
    // A DFS based function to count reachable
    // vertices from v
    private int dfsCount(Integer v, boolean[] isVisited)
    {
        // Mark the current node as visited
        isVisited[v] = true;
        int count = 1;
        // Recur for all vertices adjacent to this vertex
        for (int adj : adj[v]) {
            if (!isVisited[adj]) {
                count = count + dfsCount(adj, isVisited);
            }
        }
        return count;
    }
 
    // Driver program to test above function
    public static void main(String a[])
    {
        // Let us first create and test
        // graphs shown in above figure
        Graph g1 = new Graph(4);
        g1.addEdge(0, 1);
        g1.addEdge(0, 2);
        g1.addEdge(1, 2);
        g1.addEdge(2, 3);
        g1.printEulerTour();
 
        Graph g2 = new Graph(3);
        g2.addEdge(0, 1);
        g2.addEdge(1, 2);
        g2.addEdge(2, 0);
        g2.printEulerTour();
 
        Graph g3 = new Graph(5);
        g3.addEdge(1, 0);
        g3.addEdge(0, 2);
        g3.addEdge(2, 1);
        g3.addEdge(0, 3);
        g3.addEdge(3, 4);
        g3.addEdge(3, 2);
        g3.addEdge(3, 1);
        g3.addEdge(2, 4);
        g3.printEulerTour();
    }
}
 
// This code is contributed by Himanshu Shekhar

Python3

# Python program print Eulerian Trail in a given Eulerian or Semi-Eulerian Graph
  
from collections import defaultdict
  
#This class represents an undirected graph using adjacency list representation
class Graph:
  
    def __init__(self,vertices):
        self.V= vertices #No. of vertices
        self.graph = defaultdict(list) # default dictionary to store graph
        self.Time = 0
  
    # function to add an edge to graph
    def addEdge(self,u,v):
        self.graph[u].append(v)
        self.graph[v].append(u)
 
    # This function removes edge u-v from graph    
    def rmvEdge(self, u, v):
        for index, key in enumerate(self.graph[u]):
            if key == v:
                self.graph[u].pop(index)
        for index, key in enumerate(self.graph[v]):
            if key == u:
                self.graph[v].pop(index)
 
    # A DFS based function to count reachable vertices from v
    def DFSCount(self, v, visited):
        count = 1
        visited[v] = True
        for i in self.graph[v]:
            if visited[i] == False:
                count = count + self.DFSCount(i, visited)         
        return count
 
    # The function to check if edge u-v can be considered as next edge in
    # Euler Tour
    def isValidNextEdge(self, u, v):
        # The edge u-v is valid in one of the following two cases:
  
          #  1) If v is the only adjacent vertex of u
        if len(self.graph[u]) == 1:
            return True
        else:
            '''
             2) If there are multiple adjacents, then u-v is not a bridge
                 Do following steps to check if u-v is a bridge
  
            2.a) count of vertices reachable from u'''   
            visited =[False]*(self.V)
            count1 = self.DFSCount(u, visited)
 
            '''2.b) Remove edge (u, v) and after removing the edge, count
                vertices reachable from u'''
            self.rmvEdge(u, v)
            visited =[False]*(self.V)
            count2 = self.DFSCount(u, visited)
 
            #2.c) Add the edge back to the graph
            self.addEdge(u,v)
 
            # 2.d) If count1 is greater, then edge (u, v) is a bridge
            return False if count1 > count2 else True
 
 
    # Print Euler tour starting from vertex u
    def printEulerUtil(self, u):
        #Recur for all the vertices adjacent to this vertex
        for v in self.graph[u]:
            #If edge u-v is not removed and it's a a valid next edge
            if self.isValidNextEdge(u, v):
                print("%d-%d " %(u,v)),
                self.rmvEdge(u, v)
                self.printEulerUtil(v)
 
 
     
    '''The main function that print Eulerian Trail. It first finds an odd
   degree vertex (if there is any) and then calls printEulerUtil()
   to print the path '''
    def printEulerTour(self):
        #Find a vertex with odd degree
        u = 0
        for i in range(self.V):
            if len(self.graph[i]) %2 != 0 :
                u = i
                break
        # Print tour starting from odd vertex
        print ("\n")
        self.printEulerUtil(u)
 
# Create a graph given in the above diagram
 
g1 = Graph(4)
g1.addEdge(0, 1)
g1.addEdge(0, 2)
g1.addEdge(1, 2)
g1.addEdge(2, 3)
g1.printEulerTour()
 
 
g2 = Graph(3)
g2.addEdge(0, 1)
g2.addEdge(1, 2)
g2.addEdge(2, 0)
g2.printEulerTour()
 
g3 = Graph (5)
g3.addEdge(1, 0)
g3.addEdge(0, 2)
g3.addEdge(2, 1)
g3.addEdge(0, 3)
g3.addEdge(3, 4)
g3.addEdge(3, 2)
g3.addEdge(3, 1)
g3.addEdge(2, 4)
g3.printEulerTour()
 
 
#This code is contributed by Neelam Yadav

C#

// A C# program print Eulerian Trail
// in a given Eulerian or Semi-Eulerian Graph
using System;
using System.Collections.Generic;
 
// An Undirected graph using
// adjacency list representation
class Graph
{
    private int vertices; // No. of vertices
    private List<int>[] adj; // adjacency list
 
    // Constructor
    Graph(int numOfVertices)
    {
        // initialise vertex count
        this.vertices = numOfVertices;
 
        // initialise adjacency list
        initGraph();
    }
 
    // utility method to initialise adjacency list
    private void initGraph()
    {
        adj = new List<int>[vertices];
        for (int i = 0; i < vertices; i++) 
        {
            adj[i] = new List<int>();
        }
    }
 
    // add edge u-v
    private void addEdge(int u, int v)
    {
        adj[u].Add(v);
        adj[v].Add(u);
    }
 
    // This function removes edge u-v from graph.
    private void removeEdge(int u, int v)
    {
        adj[u].Remove(v);
        adj[v].Remove(u);
    }
 
    /* The main function that print Eulerian Trail. 
    It first finds an odd degree vertex (if there 
    is any) and then calls printEulerUtil() to
    print the path */
    private void printEulerTour()
    {
        // Find a vertex with odd degree
        int u = 0;
        for (int i = 0; i < vertices; i++)
        {
            if (adj[i].Count % 2 == 1)
            {
                u = i;
                break;
            }
        }
         
        // Print tour starting from oddv
        printEulerUtil(u);
        Console.WriteLine();
    }
 
    // Print Euler tour starting from vertex u
    private void printEulerUtil(int u)
    {
        // Recur for all the vertices
        // adjacent to this vertex
        for (int i = 0; i < adj[u].Count; i++)
        {
            int v = adj[u][i];
             
            // If edge u-v is a valid next edge
            if (isValidNextEdge(u, v)) 
            {
                Console.Write(u + "-" + v + " ");
                 
                // This edge is used so remove it now
                removeEdge(u, v); 
                printEulerUtil(v);
            }
        }
    }
 
    // The function to check if edge u-v can be
    // considered as next edge in Euler Tout
    private bool isValidNextEdge(int u, int v)
    {
        // The edge u-v is valid in one of the
        // following two cases:
 
        // 1) If v is the only adjacent vertex of u 
        // ie size of adjacent vertex list is 1
        if (adj[u].Count == 1) 
        {
            return true;
        }
 
        // 2) If there are multiple adjacents, then
        // u-v is not a bridge Do following steps 
        // to check if u-v is a bridge
        // 2.a) count of vertices reachable from u
        bool[] isVisited = new bool[this.vertices];
        int count1 = dfsCount(u, isVisited);
 
        // 2.b) Remove edge (u, v) and after removing
        // the edge, count vertices reachable from u
        removeEdge(u, v);
        isVisited = new bool[this.vertices];
        int count2 = dfsCount(u, isVisited);
 
        // 2.c) Add the edge back to the graph
        addEdge(u, v);
        return (count1 > count2) ? false : true;
    }
 
    // A DFS based function to count reachable
    // vertices from v
    private int dfsCount(int v, bool[] isVisited)
    {
        // Mark the current node as visited
        isVisited[v] = true;
        int count = 1;
         
        // Recur for all vertices adjacent
        // to this vertex
        foreach(int i in adj[v])
        {
            if (!isVisited[i])
            {
                count = count + dfsCount(i, isVisited);
            }
        }
        return count;
    }
 
    // Driver Code
    public static void Main(String []a)
    {
        // Let us first create and test
        // graphs shown in above figure
        Graph g1 = new Graph(4);
        g1.addEdge(0, 1);
        g1.addEdge(0, 2);
        g1.addEdge(1, 2);
        g1.addEdge(2, 3);
        g1.printEulerTour();
 
        Graph g2 = new Graph(3);
        g2.addEdge(0, 1);
        g2.addEdge(1, 2);
        g2.addEdge(2, 0);
        g2.printEulerTour();
 
        Graph g3 = new Graph(5);
        g3.addEdge(1, 0);
        g3.addEdge(0, 2);
        g3.addEdge(2, 1);
        g3.addEdge(0, 3);
        g3.addEdge(3, 4);
        g3.addEdge(3, 2);
        g3.addEdge(3, 1);
        g3.addEdge(2, 4);
        g3.printEulerTour();
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

// A Javascript program print Eulerian Trail in a given Eulerian or
// Semi-Eulerian Graph
 
      // A class that represents an undirected graph
      class Graph {
        constructor(V) {
          this.V = V; // No. of vertices
          //A dynamic array of adjacency lists
          this.adj = Array.from(Array(V), () => new Array());
        }
 
        // functions to add edge
        addEdge(u, v) {
          this.adj[u].push(v);
          this.adj[v].push(u);
        }
 
        // This function removes edge u-v from graph.
        // It removes the edge by replacing adjacent
        // vertex value with -1.
        rmvEdge(u, v) {
          // Find v in adjacency list of u and replace
          // it with -1
 
          for (let i = 0; i < this.adj[u].length; i++) {
            if (this.adj[u][i] == v) {
              this.adj[u][i] = -1;
              break;
            }
          }
 
          // Find u in adjacency list of v and replace
          // it with -1
 
          for (let i = 0; i < this.adj[v].length; i++) {
            if (this.adj[v][i] == u) {
              this.adj[v][i] = -1;
              break;
            }
          }
        }
 
        // Methods to print Eulerian tour
        printEulerTour() {
          // Find a vertex with odd degree
          let u = 0;
 
          for (let i = 0; i < this.V; i++)
            if (this.adj[i].length & 1) {
              u = i;
              break;
            }
 
          // Print tour starting from oddv
          this.printEulerUtil(u);
          console.log("   ");
        }
 
        //Print Euler tour starting from vertex u
        printEulerUtil(u) {
          {
            // Recur for all the vertices adjacent to
            // this vertex
 
            for (let j in this.adj[u]) {
              let v = this.adj[u][j];
 
              // If edge u-v is not removed and it's a
              // valid next edge
              if (v != -1 && this.isValidNextEdge(u, v)) {
                console.log(u + "-" + v);
                this.rmvEdge(u, v);
                this.printEulerUtil(v);
              }
            }
          }
        }
 
        // This function returns count of vertices
        // reachable from v. It does DFS
        DFSCount(v, visited) {
          // Mark the current node as visited
          visited[v] = true;
          let count = 1;
 
          // Recur for all vertices adjacent to this vertex
 
          for (let j in this.adj[v]) {
            let i = this.adj[v][j];
            if (i != -1 && !visited[i]) count += this.DFSCount(i, visited);
          }
          return count;
        }
 
        // The function to check if edge u-v can be considered
        // as next edge in Euler Tout
        isValidNextEdge(u, v) {
          // The edge u-v is valid in one of the following
          // two cases:
 
          // 1) If v is the only adjacent vertex of u
          let count = 0; // To store count of adjacent vertices
 
          for (let j in this.adj[u]) {
            let i = this.adj[u][j];
            if (i != -1) count++;
          }
          if (count == 1) return true;
 
          // 2) If there are multiple adjacents, then u-v
          //    is not a bridge
          // Do following steps to check if u-v is a bridge
 
          // 2.a) count of vertices reachable from u
          let visited = new Array(this.V);
          visited.fill(false);
          let count1 = this.DFSCount(u, visited);
 
          // 2.b) Remove edge (u, v) and after removing
          // the edge, count vertices reachable from u
          this.rmvEdge(u, v);
          visited.fill(false);
          let count2 = this.DFSCount(u, visited);
 
          // 2.c) Add the edge back to the graph
          this.addEdge(u, v);
 
          // 2.d) If count1 is greater, then edge (u, v)
          // is a bridge
          return count1 > count2 ? false : true;
        }
      }
 
      // Driver program to test above function
 
      // Let us first create and test graphs shown in above
      // figure
      let g1 = new Graph(4);
      g1.addEdge(0, 1);
      g1.addEdge(0, 2);
      g1.addEdge(1, 2);
      g1.addEdge(2, 3);
      g1.printEulerTour();
 
      let g2 = new Graph(3);
      g2.addEdge(0, 1);
      g2.addEdge(1, 2);
      g2.addEdge(2, 0);
      g2.printEulerTour();
 
      let g3 = new Graph(5);
      g3.addEdge(1, 0);
      g3.addEdge(0, 2);
      g3.addEdge(2, 1);
      g3.addEdge(0, 3);
      g3.addEdge(3, 4);
      g3.addEdge(3, 2);
      g3.addEdge(3, 1);
      g3.addEdge(2, 4);
      g3.printEulerTour();

Output

2-0  0-1  1-2  2-3  
0-1  1-2  2-0  
0-1  1-2  2-0  0-3  3-4  4-2  2-3  3-1

Note that the above code modifies the given graph, we can create a copy of the graph if we don’t want the given graph to be modified.

Time Complexity: The time complexity of the above implementation is O ((V+E)²). The function printEulerUtil() is like DFS and it calls isValidNextEdge() which also does DFS two times. The time complexity of DFS for adjacency list representation is O(V+E). Therefore overall time complexity is O((V+E)*(V+E)) which can be written as O(E²) for a connected graph.
There are better algorithms to print Euler tour, Hierholzer’s Algorithm finds in O(V+E) time.
Space complexity: O(V+E),the space complexity of the above program is O(V+E) as we are using an adjacency list to represent the graph. The size of the adjacency list is equal to the number of vertices plus the number of edges in the graph.