I've read in different sources that the amortized cost of a red-black tree rebalancing is constant (at least during the tree creation using only insertions).

How can it be proven?

  • 1
    $\begingroup$ What do these "different" sources have to say about that? What are these sources, anyway? $\endgroup$
    – Raphael
    Feb 3, 2016 at 23:01
  • $\begingroup$ @Raphael: The first ten blue links on google.com for "amortized" "red-black" are some examples, but about half of them say exactly how it can be proved. I'm not sure what else the poster thinks is left out of them. $\endgroup$
    – jbapple
    Feb 4, 2016 at 1:25

1 Answer 1


If you look at the rules for what happens in a red/black tree insertion, you can see that the fixup rules for maintaining the red/black invariants only propagate upward if the newly-inserted node becomes the child of a red node with a red sibling.

Let's call a "fissionable" group in the red/black tree a black node with two red children. If you do an insertion into a fissionable group, the fissionable group will be recolored and rotated in a way that ends up eliminating that fissionable group, but which propagates the insertion fixup to a higher level in the tree. If you set the potential function of a red/black tree to be the number of fissionable groups in the tree, then the true cost of doing an insertion is 1 + r, where r is the number of groups destroyed, and the number of new fissionable groups created is at most one (you should check this), so the amortized cost of a fixup is O(1) because the potential drop is r.

It's a lot easier to think about this by looking at the isometry between red/black trees and 2-3-4 trees. A fissionable group in a red/black tree corresponds to a 4-node in a 2-3-4 tree, and if you understand 2-3-4 tree insertions it's pretty easy to see why the amortized cost of an insertion into a red/black tree is O(1).


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