Detect cycle in an undirected graph

Last Updated : 07 Apr, 2025

Given an undirected graph, the task is to check if there is a cycle in the given graph.

Examples:

Input: V = 4, edges[][]= [[0, 1], [0, 2], [1, 2], [2, 3]]
Undirected Graph with 4 vertices and 4 edges
Output: true
Explanation: The diagram clearly shows a cycle 0 → 2 → 1 → 0
Input: V = 4, edges[][] = [[0, 1], [1, 2], [2, 3]]
Undirected graph with 4 vertices and 3 edges
Output: false
Explanation: There is no cycle in the given graph.

Table of Content

Using Breadth First Search - O(V+E) Time and O(V) Space

BFS is useful for cycle detection in an undirected graph because it explores level by level, ensuring that each node is visited in the shortest possible way. It efficiently detects cycles using a visited array and a queue while avoiding unnecessary recursive calls, making it more memory-efficient than DFS for large graphs.

During BFS traversal, we maintain a visited array and a queue. We process nodes by popping them one by one from the queue, marking them as visited, and pushing their unvisited adjacent nodes into the queue. A cycle is detected if we encounter a node that has already been visited before being dequeued, meaning it has been reached through a different path. This approach ensures that we efficiently detect cycles while maintaining optimal performance.

Please refer Detect cycle in an undirected graph using BFS for complete implementation.

Using Depth First Search - O(V+E) Time and O(V) Space

Depth First Traversal can be used to detect a cycle in an undirected Graph. If we encounter a visited vertex again, then we say, there is a cycle. But there is a catch in this algorithm, we need to make sure that we do not consider every edge as a cycle because in an undirected graph, an edge from 1 to 2 also means an edge from 2 to 1. To handle this, we keep track of the parent node (the node from which we came to the current node) in the DFS traversal and ignore the parent node from the visited condition.

Follow the below steps to implement the above approach:

Iterate over all the nodes of the graph and Keep a visited array visited[] to track the visited nodes.
If the current node is not visited, run a Depth First Traversal on the given subgraph connected to the current node and pass the parent of the current node as -1. Recursively, perform the following steps:
- Set visited[root] as 1.
- Iterate over all adjacent nodes of the current node in the adjacency list
  - If it is not visited then run DFS on that node and return true if it returns true.
  - Else if the adjacent node is visited and not the parent of the current node then return true.
- Return false.

Illustration:

Below is the graph showing how to detect cycle in a graph using DFS:

Below is the implementation of the above approach:

C++

// A C++ Program to detect cycle in an undirected graph #include <bits/stdc++.h> using namespace std;  bool isCycleUtil(int v, vector<vector<int>> &adj, vector<bool> &visited, int parent) {     // Mark the current node as visited     visited[v] = true;      // Recur for all the vertices adjacent to this vertex     for (int i : adj[v])     {         // If an adjacent vertex is not visited, then recur for that adjacent         if (!visited[i])         {             if (isCycleUtil(i, adj, visited, v))                 return true;         }         // If an adjacent vertex is visited and is not parent of current vertex,         // then there exists a cycle in the graph.         else if (i != parent)             return true;     }      return false; } vector<vector<int>> constructadj(int V, vector<vector<int>> &edges){          vector<vector<int>> adj(V);     for (auto it : edges)     {         adj[it[0]].push_back(it[1]);         adj[it[1]].push_back(it[0]);     }          return adj; } // Returns true if the graph contains a cycle, else false. bool isCycle(int V, vector<vector<int>> &edges) {          vector<vector<int>> adj = constructadj(V,edges);     // Mark all the vertices as not visited     vector<bool> visited(V, false);      for (int u = 0; u < V; u++)     {         if (!visited[u])         {             if (isCycleUtil(u, adj, visited, -1))                 return true;         }     }      return false; }  int main() {     int V = 5;     vector<vector<int>> edges = {{0, 1}, {0, 2}, {0, 3}, {1, 2}, {3, 4}};      if (isCycle(V, edges))     {         cout << "true" << endl;     }     else     {         cout << "false" << endl;     }      return 0; }

Java

import java.util.*;  class GfG {     // Helper function to check cycle using DFS     static boolean isCycleUtil(int v, List<Integer>[] adj,                                boolean[] visited,                                int parent)     {         visited[v] = true;         // If an adjacent vertex is not visited,         // then recur for that adjacent         for (int i : adj[v]) {             if (!visited[i]) {                 if (isCycleUtil(i, adj, visited, v))                     return true;             }             // If an adjacent vertex is visited and             // is not parent of current vertex,             // then there exists a cycle in the graph.             else if (i != parent) {                 return true;             }         }         return false;     }     static  List<Integer>[] constructadj(int V, int [][] edges){                  List<Integer>[] adj = new ArrayList[V];          for (int i = 0; i < V; i++) {             adj[i] = new ArrayList<>();         }                  return adj;     }      // Function to check if graph contains a cycle     static boolean isCycle(int V, int[][] edges)     {         List<Integer> [] adj = constructadj(V,edges);                  for (int[] edge : edges) {             adj[edge[0]].add(edge[1]);             adj[edge[1]].add(edge[0]);         }          boolean[] visited = new boolean[V];         // Call the recursive helper function         // to detect cycle in different DFS trees         for (int u = 0; u < V; u++) {             if (!visited[u]) {                 if (isCycleUtil(u, adj, visited, -1))                     return true;             }         }         return false;     }      public static void main(String[] args)     {         int V = 5;         int[][] edges = {             {0, 1}, {0, 2}, {0, 3}, {1, 2}, {3, 4}         };          if (isCycle(V, edges)) {             System.out.println("true");         }         else {             System.out.println("false");         }     } }

Python

# Helper function to check cycle using DFS def isCycleUtil(v, adj, visited, parent):     visited[v] = True      for i in adj[v]:         if not visited[i]:             if isCycleUtil(i, adj, visited, v):                 return True         elif i != parent:             return True     return False  def constructadj(V, edges):     adj = [[] for _ in range(V)]  # Initialize adjacency list      for edge in edges:         u, v = edge         adj[u].append(v)         adj[v].append(u)          return adj  # Function to check if graph contains a cycle def isCycle(V, edges):          adj = constructadj(V,edges)     visited = [False] * V      for u in range(V):         if not visited[u]:             if isCycleUtil(u, adj, visited, -1):                 return True     return False   # Driver Code if __name__ == "__main__":     V = 5     edges = [(0, 1), (0, 2), (0, 3), (1, 2), (3, 4)]      if isCycle(V, edges):         print("true")     else:         print("false")

using System; using System.Collections.Generic;  class CycleDetection {     // Helper function to check cycle using DFS     static bool IsCycleUtil(int v, List<int>[] adj,                             bool[] visited, int parent)     {         visited[v] = true;          foreach(int i in adj[v])         {             if (!visited[i]) {                 if (IsCycleUtil(i, adj, visited, v))                     return true;             }             else if (i != parent) {                 return true;             }         }         return false;     }     static List<int>[] constructadj(int V, int [,] edges){         List<int>[] adj = new List<int>[ V ];          for (int i = 0; i < V; i++) {             adj[i] = new List<int>();         }                  return adj;     }     // Function to check if graph contains a cycle     static bool IsCycle(int V, int[, ] edges)     {            List<int>[] adj = constructadj(V,edges);         for (int i = 0; i < edges.GetLength(0); i++) {             int u = edges[i, 0], v = edges[i, 1];             adj[u].Add(v);             adj[v].Add(u);         }          bool[] visited = new bool[V];          for (int u = 0; u < V; u++) {             if (!visited[u]) {                 if (IsCycleUtil(u, adj, visited, -1))                     return true;             }         }         return false;     }      public static void Main()     {         int V = 5;         int[, ] edges = {             { 0, 1 }, { 0, 2 }, { 0, 3 }, { 1, 2 }, { 3, 4 }         };          if (IsCycle(V, edges)) {             Console.WriteLine("true");         }         else {             Console.WriteLine("false");         }     } }

JavaScript

// Helper function to check cycle using DFS function isCycleUtil(v, adj, visited, parent) {     visited[v] = true;      for (let i of adj[v]) {         if (!visited[i]) {             if (isCycleUtil(i, adj, visited, v)) {                 return true;             }         }         else if (i !== parent) {             return true;         }     }     return false; }  function constructadj(V, edges){     let adj = Array.from({length : V}, () => []);      // Build the adjacency list     for (let edge of edges) {         let [u, v] = edge;         adj[u].push(v);         adj[v].push(u);     }     return adj; } // Function to check if graph contains a cycle function isCycle(V, edges) {     let adj = constructadj(V,edges);      let visited = new Array(V).fill(false);      // Check each node     for (let u = 0; u < V; u++) {         if (!visited[u]) {             if (isCycleUtil(u, adj, visited, -1)) {                 return true;             }         }     }     return false; }  // Driver Code const V = 5; const edges =     [[0, 1], [0, 2], [0, 3], [1, 2], [3, 4]];  if (isCycle(V, edges)) {     console.log("true"); } else {     console.log("false"); }

Output

true

Time Complexity: O(V+E) because DFS visits each vertex once (O(V)) and traverses all edges once (O(E))
Auxiliary space: O(V) for the visited array and O(V) for the recursive call stack.

We do not count the adjacency list in auxiliary space as it is necessary for representing the input graph.

Related Articles:

Detect cycle in an undirected graph

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Detect cycle in an undirected graph

Using Breadth First Search - O(V+E) Time and O(V) Space

Using Depth First Search - O(V+E) Time and O(V) Space

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