Order of insertion for worst case black height of a red black tree

Question

Lets say we are dealing with the keys 1-15. To get the worst case performance of a regular BST, you would insert the keys in ascending or descending order as follows:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

Then the BST would essentially become a linked list.

For best case of a BST you would insert the keys in the following order, they are arranged in such a way that the next key inserted is half of the total range to be inserted, so the first is 15/2 = 8, then 8/2 = 4 etc...

8, 4, 12, 2, 6, 10, 14, 1, 3, 5, 7, 9, 11, 13, 15

Then the BST would be a well balanced tree with optimal height 3.

The best case for a red black tree can also be constructed with the best case from a BST. But how do we construct the worst case for a red black tree? Is it the same as the worst case for a BST? Is there a specific pattern that will yield the worst case?

Answer 1

You are looking for a skinny tree, right? This can be produced by inserting [1 ... , 2^(n+1)-2] in reverse order.

Answer 2

You won't be able to. A Red-Black Tree keeps itself "bushy", so it would rotate to fix the imbalance. The length of your above worst case for a Red-Black Tree is limited to two elements, but that's still not a "bad" case; it's what's expected, as lg(2) = 1, and you have 1 layer past the root with two elements. As soon as you add the third element, you get this:

B                   B
 \                 / \
  R       =>      R   R
   \
    R