问题
Has anyone seen an implementation of the STL where stl::set is not implemented as a red-black tree?
The reason I ask is that, in my experiments, B-2B trees outperform stl::set (and other red-black tree implementations) by a factor of 2 to 4 depending on the value of B. I'm curious if there is a compelling reason to use red-black trees when there appear to be faster data structures available.
回答1:
Some folks over at Google actually built a B-tree based implementation of the C++ standard library containers. They seem to have much better performance than standard binary tree implementations.
There is a catch, though. The C++ standard guarantees that deleting an element from a map or set only invalidates other iterators pointing to the same element in the map or set. With the B-tree based implementation, due to node splits and consolidations, the erase member functions on these new structures may invalidate iterators to other elements in the tree. As a result, these implementations aren't perfect replacements for the standard implementations and couldn't be used in a conformant implementation.
Hope this helps!
回答2:
There is at least one implementation based on AVL trees instead of red-black trees.
I haven't tried to verify conformance of this implementation, but at least (unlike a B-tree based implementation) it at least could be written to conform perfectly to the requirements of the standard.
来源：https://stackoverflow.com/questions/26550276/does-any-stlset-implementation-not-use-a-red-black-tree