Python Program for Breadth First Search or BFS for a Graph
Breadth First Traversal (or Search) for a graph is similar to Breadth First Traversal of a tree (See method 2 of this post).
The only catch here is, that unlike trees, graphs may contain cycles, so we may come to the same node again. To avoid processing a node more than once, we use a boolean visited array. For simplicity, it is assumed that all vertices are reachable from the starting vertex. For example, in the following graph, we start traversal from vertex 2. When we come to vertex 0, we look for all adjacent vertices of it. 2 is also an adjacent vertex of 0. If we don’t mark visited vertices, then 2 will be processed again and it will become a non-terminating process. A Breadth First Traversal of the following graph is 2, 0, 3, 1.
Following are the implementations of simple Breadth First Traversal from a given source.
The implementation uses adjacency list representation of graphs. STL\’s list container is used to store lists of adjacent nodes and queue of nodes needed for BFS traversal.
# Python3 Program to print BFS traversal
# from a given source vertex. BFS(int s)
# traverses vertices reachable from s.
from collections import defaultdict
# This class represents a directed graph
# using adjacency list representation
class Graph:
# Constructor
def __init__(self):
# Default dictionary to store graph
self.graph = defaultdict(list)
# Function to add an edge to graph
def addEdge(self, u, v):
self.graph[u].append(v)
# Function to print a BFS of graph
def BFS(self, s):
# Mark all the vertices as not visited
visited = [False] * (max(self.graph) + 1)
# Create a queue for BFS
queue = []
# Mark the source node as
# visited and enqueue it
queue.append(s)
visited[s] = True
while queue:
# Dequeue a vertex from
# queue and print it
s = queue.pop(0)
print(s, end=" ")
# Get all adjacent vertices of the
# dequeued vertex s.
# If an adjacent has not been visited,
# then mark it visited and enqueue it
for i in self.graph[s]:
if not visited[i]:
queue.append(i)
visited[i] = True
# Driver code
if __name__ == '__main__':
# Create a graph given in
# the above diagram
g = Graph()
g.addEdge(0, 1)
g.addEdge(0, 2)
g.addEdge(1, 2)
g.addEdge(2, 0)
g.addEdge(2, 3)
g.addEdge(3, 3)
print("Following is Breadth First Traversal"
" (starting from vertex 2)")
g.BFS(2)
# This code is contributed by Neelam Yadav
# This code is modified by Susobhan Akhuli
Output
Following is Breadth First Traversal (starting from vertex 2) 2 0 3 1
Time Complexity : O(V+E), where V is the number of vertices in graph and E is the number of edges
Auxiliary Space: O(V)
Please refer complete article on Breadth First Search or BFS for a Graph for more details!