I liked qiao's first answer very much!
The only thing missing here is to mark the vertexes as visited.
Why we need to do it?
Lets imagine that there is another node number 13 connected from node 11. Now our goal is to find node 13.
After a little bit of a run the queue will look like this:
[[1, 2, 6], [1, 3, 10], [1, 4, 7], [1, 4, 8], [1, 2, 5, 9], [1, 2, 5, 10]]
Note that there are TWO paths with node number 10 at the end.
Which means that the paths from node number 10 will be checked twice. In this case it doesn't look so bad because node number 10 doesn't have any children.. But it could be really bad (even here we will check that node twice for no reason..)
Node number 13 isn't in those paths so the program won't return before reaching to the second path with node number 10 at the end..And we will recheck it..
All we are missing is a set to mark the visited nodes and not to check them again..
This is qiao's code after the modification:
graph = {
1: [2, 3, 4],
2: [5, 6],
3: [10],
4: [7, 8],
5: [9, 10],
7: [11, 12],
11: [13]
}
def bfs(graph_to_search, start, end):
queue = [[start]]
visited = set()
while queue:
# Gets the first path in the queue
path = queue.pop(0)
# Gets the last node in the path
vertex = path[-1]
# Checks if we got to the end
if vertex == end:
return path
# We check if the current node is already in the visited nodes set in order not to recheck it
elif vertex not in visited:
# enumerate all adjacent nodes, construct a new path and push it into the queue
for current_neighbour in graph_to_search.get(vertex, []):
new_path = list(path)
new_path.append(current_neighbour)
queue.append(new_path)
# Mark the vertex as visited
visited.add(vertex)
print bfs(graph, 1, 13)
The output of the program will be:
[1, 4, 7, 11, 13]
Without the unneccecery rechecks..