Timeline for Depth-first search in Python
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 1, 2017 at 8:53 | comment | added | BBSysDyn | You are using set to keep track of visit order but set() is not guaranteed to give correct order. visited should be a list. | |
| S Feb 20, 2015 at 12:14 | history | suggested | CommunityBot | CC BY-SA 3.0 |
bfs typo; changed to complete term
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| Feb 20, 2015 at 11:50 | review | Suggested edits | |||
| S Feb 20, 2015 at 12:14 | |||||
| Feb 5, 2015 at 3:48 | vote | accept | meto | ||
| Jan 26, 2015 at 12:57 | comment | added | sapi | @JanneKarila Good point; I was picturing a top-down tree (where that won't occur for a DFS), but of course that's a very specific case. | |
| Jan 26, 2015 at 11:05 | comment | added | Janne Karila |
@sapi A node would be pushed multiple times to stack if it can be reached from multiple nodes that happen to be visited before it.
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| Jan 26, 2015 at 10:02 | comment | added | f.rodrigues | Makes sense. Seems redundant. The code ain't mine tho, the link provide may have some info on that. | |
| Jan 26, 2015 at 9:47 | comment | added | sapi |
Unless I'm missing something, if you only ever add unvisited nodes to the stack, you shouldn't need the if vertex not in visited check (presuming a node cannot have two links to the same child). Which you want would depend on whether set subtraction or contains is faster (I would guess the latter).
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| Jan 26, 2015 at 4:53 | history | answered | f.rodrigues | CC BY-SA 3.0 |