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patternpythonMinor

Depth and breadth first search

Submitted by: @import:stackexchange-codereview··
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depthsearchbreadthfirstand

Problem

I am practicing interview questions, I wrote this code to perform BFS and DFS on a graph in Python.

How can it be optimized, and how can it be made more readable?

class Graph:
    def __init__(self):
        self.graph = defaultdict(list)

    def add_edge(self, u, v):
        self.graph[u].append(v)

    def dfs(self, node):
        visited = [False for i in range(len(self.graph))]
        stack = []
        stack.append(node)
        visited[node] = True
        while stack:
            node = stack.pop()
            print node
            for i in self.graph[node]:
                if visited[i] == False:
                    visited[i] = True
                    stack.append(i)

     def Bfs(self, node):
        visited = [False for i in range(len(self.graph))]
        queue = []
        queue.append(node)
        visited[node] = True
        while queue:
            node = queue.pop(0)
            print node
            for i in self.graph[node]:
                if not visited[i]:
                    queue.append(i)
                    visited[i] = True

Solution

Small syntactical changes

Bfs should at least be bfs. Some would probably argue that you should spell out depth_first_search and breadth_first_search entirely. I personally like small names, but regardless, the capital B should not be there.

You wrote this one time:

if visited[i] == False:


And used the correct idiom the second time:

if not visited[i]:


Performance

According to here, generating a list with multiplication is the fastest way, so change both occurrences of:

visited = [False for i in range(len(self.graph))]


to:

visited = [False] * len(self.graph)

Code Snippets

if visited[i] == False:
if not visited[i]:
visited = [False for i in range(len(self.graph))]
visited = [False] * len(self.graph)

Context

StackExchange Code Review Q#159237, answer score: 3

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