When you switch to a red-black tree, is there any reason to choose one shape over another?

Question

I have a library of linked lists / binary tree methods for use when standard containers are not suitable - for example. when there are different types of nodes, or when I need to convert from a binary tree to a list and vice versa. It includes red-black wood processing.

One of the methods is converted from a double list to a perfectly balanced simple binary tree in O(n)time (given that the number of elements is known in advance). The algorithm is known as "folding" - this is the second half of the binary tree balancing algorithm that was once published by Dr. Dobbs. The steps are basically ...

• Given the size of the tree, determine the size of the left and right subtrees

• Count left subtree

• Put a node from the list to be used as root

• Recurse for the correct subtree

• Link the subtrees to the root

I also have a similar method that creates a red-black tree. The principle is the same, but recursion tracks node heights - zero vertices are created in red, all the rest are black. The initial height calculation is based on the highest bit set in tree size and is split so that a perfectly balanced tree of size (2^n)-1has only black nodes (recursion only drops to a height of one).

The point here is that I only have red nodes at the sheet level, and at most half the nodes are red.

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