Timeline for Implementing an algorithm that walks the DOM without recursion
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2017 at 10:51 | comment | added | mike rodent |
PPS also there is a slicker way to do the reversing and adding of the child nodes in ES6: stack.push( ... Array.from( node.childNodes ).reverse() ); - no reassignment of stack, so stack can be const.
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| Sep 16, 2017 at 16:35 | comment | added | mike rodent | Very nice... please see my slight variation on this theme, in which you can get the depth of each node as you iterate... | |
| Jul 25, 2015 at 1:09 | comment | added | Andrew |
@o11c I don't know about a better way, but some may find this easier to read: for (var i = node.childNodes.length-1; i >= 0; i--) { stack.push(node.childNodes[i]); }
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| Jul 25, 2015 at 1:01 | history | edited | Andrew | CC BY-SA 3.0 |
Changed the word 'simple' to 'concise' to be more clear.
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| Jul 25, 2015 at 1:00 | comment | added | Andrew | @Hosch250 which original version are you referring to? The one with recursion? The original poster was asking for a concise solution without recursion: that's what I was going for. | |
| Jul 24, 2015 at 1:42 | comment | added | o11c | I do wonder if it's possible to do the append-all-reversed in a better way. I'm not nearly familiar enough with javascript to know. | |
| Jul 24, 2015 at 1:35 | comment | added | o11c | Ooh, clever. I hadn't thought of this as a DFS problem. You deserve more upvotes! | |
| Jul 24, 2015 at 0:31 | review | Late answers | |||
| Jul 24, 2015 at 0:42 | |||||
| Jul 24, 2015 at 0:26 | comment | added | user34073 | Can you explain why your alternative version is better than the original version? | |
| Jul 24, 2015 at 0:22 | review | First posts | |||
| Jul 24, 2015 at 0:26 | |||||
| Jul 24, 2015 at 0:13 | history | answered | Andrew | CC BY-SA 3.0 |