Graph Traversal
Breadth-First Search or BFS
Breadth-First Traversal for a graph is similar to Breadth-First Traversal of a tree. The only catch here is, 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.
# 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 a adjacent
# has not been visited, then mark it
# visited and enqueue it
for i in self.graph[s]:
if visited[i] == False:
queue.append(i)
visited[i] = True
# Driver code
# 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)
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 the graph and E is the number of edges in the graph.
Depth First Search or DFS
Depth First Traversal for a graph is similar to Depth First Traversal of a tree. The only catch here is, unlike trees, graphs may contain cycles, a node may be visited twice. To avoid processing a node more than once, use a boolean visited array.
Algorithm:
- Create a recursive function that takes the index of the node and a visited array.
- Mark the current node as visited and print the node.
- Traverse all the adjacent and unmarked nodes and call the recursive function with the index of the adjacent node.
# Python3 program to print DFS traversal
# from a given graph
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)
# A function used by DFS
def DFSUtil(self, v, visited):
# Mark the current node as visited
# and print it
visited.add(v)
print(v, end=' ')
# Recur for all the vertices
# adjacent to this vertex
for neighbour in self.graph[v]:
if neighbour not in visited:
self.DFSUtil(neighbour, visited)
# The function to do DFS traversal. It uses
# recursive DFSUtil()
def DFS(self, v):
# Create a set to store visited vertices
visited = set()
# Call the recursive helper function
# to print DFS traversal
self.DFSUtil(v, visited)
# Driver code
# 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 DFS from (starting from vertex 2)")
g.DFS(2)
Output
Following is DFS from (starting from vertex 2) 2 0 1 3
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