current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Code Review Stack Exchange is a question and answer site for peer programmer code reviews. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
up vote 7 down vote favorite 
def bfs(graph, root):
    visited, queue = [], [root]
    while queue:
        vertex = queue.pop(0)
        for w in graph[vertex]:
            if w not in visited:
                visited.append(w)
                queue.append(w)

graph = {0: [1, 2], 1: [2], 2: []}
bfs(graph, 0)

Does this look like a correct implementation of BFS in Python 3?

share|improve this question
up vote 10 down vote accepted
  • sets perform containing checks (w in visited) \$O(1)\$ rather than \$O(n)\$ for lists.
  • collections.deque are better than lists for poping elements at the front (popleft).
  • you should put your example code under an if __name__ == '__main__' clause.
  • w as a variable name does not convey meaning, you should try to come up with something more explicit.
import collections


def breadth_first_search(graph, root): 
    visited, queue = set(), collections.deque([root])
    while queue: 
        vertex = queue.popleft()
        for neighbour in graph[vertex]: 
            if neighbour not in visited: 
                visited.add(neighbour) 
                queue.append(neighbour) 


if __name__ == '__main__':
    graph = {0: [1, 2], 1: [2], 2: []} 
    breadth_first_search(graph, 0)
share|improve this answer
    
thanks for this critique. – Apollo Jul 18 at 17:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.