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

7
votes

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?

2
binary-search-treered-black-treeinsertion-order
Hey this is a great question, I think. And I am particularly interested in knowing the answer. Perhaps people at cstheory stackexchange can help here. So if you can post it there? – sukunrt
I came across this paper: citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.46.1458 – sukunrt
Could you clarify half of the total range to be inserted? Just curious, and couldn't figure it out – keyser
The worst case is rather good. Wikipedia says: "The balancing of the tree is not perfect, but it is good enough to allow it to guarantee searching in O(log n) time, where n is the total number of elements in the tree." and: "[...] valuable in [...] real-time applications [...] and the Completely Fair Scheduler used in current Linux kernels and epoll system call implementation uses red–black trees." – nalply

2 Answers

2
votes

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

0
votes

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