Quote (SilverMice @ Jan 17 2013 10:03pm)
Thanks, I've also looked at it but I'd prefer the idea behind it instead of the code.
@
Quote (Azrad @ Jan 17 2013 10:12pm)
well the number of bits required to number a 'row' in the tree, is equal to the number of rows above it, for example:
here the final row requires 3 bits to enumerate the row, so there are 3 rows above it
http://s1.postimage.org/bxex2i5m7/Untitled.png
This is an interesting approach - how would you label a node that is not existent? Because what you drew is balanced on both sides; but, what if a tree has many branches with only 1 node that has 3 bits - that would mean I'd have to transcend the tree to find it :o.