blog/ib002/08-rb-trees/rules.md

title	description	tags
On the rules of the red-black tree	Shower thoughts on the rules of the red-black tree.	red-black trees balanced trees

Introduction

Have you ever thought about the red-black tree rules in more depth? Why are they formulated the way they are? How come they keep the tree balanced? Let's go through each of the red-black tree rules and try to change, break and contemplate about them.

We expect that you are familiar with the following set of the rules¹:

1. Every node is either red or black.
2. The root is black.
3. Every leaf (nil) is black.
4. If a node is red, then both its children are black.
5. For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.

Each section will go into reasonable details of each rule.

1ª Every node is either red or black.

OK… This one is very simple. It is just a definition and is used in all other rules. Not much to talk about here. Or is there?

Do I really need the nodes to be explicitly colored?

The answer is no. Balancing of the red-black trees is “enforced” by the 4th and 5th rule in the enumeration above. There are many ways you can avoid using colors.

Black height

We mentioned the 4th and 5th rule and that it enforces the balancing. What does it mean for us?

Well, we definitely do not have to use the colors, which even as a boolean flag would take at least 1 byte of space (and usually even more), cause… well, it is easier for the CPU to work with words rather than single bits.

We could use the black height, couldn't we? It would mean more memory used, cause it should be ideally big and unsigned. Can we tell the color of a node from the black height? Of course we can, if my child has the same black height as I do, it means that there was no black node added on the path between us and therefore my child would be colored red.

Example of a red-black tree that keeps count of black nodes on paths to the leaves follows:

We mark the black heights in superscript. You can see that all leaves have the black height equal to 1. Let's take a look at some of the interesting cases:

• If we take a look at the node with \text{key} = 9, we can see that it is coloured red and its black height is 1, because it is a leaf.

  Let's look at its parent (node with \text{key} = 8). On its left side it has nil and on its right side the 9. And its black height is still 1, cause except for the nil leaves, there are no other black nodes.

  We can clearly see that if a node has the same black height as its parent, it is a red node.

• Now let's take a look at the root with \text{key} = 3. It has a black height of 3. Both of its children are black nodes and have black height of 2.

  We can see that if a node has its height 1 lower than its parent, it is a black node.

  The reasoning behind it is rather simple, we count the black nodes all the way to the leaves, therefore if my parent has a higher black height, it means that on the path from me to my parent there is a black node, but the only node added is me, therefore I must be black.

Isomorphic trees

One of the other ways to avoid using color is storing the red-black tree in some isomorphic tree. The structure of 2-3-4 tree allows us to avoid using the color completely. This is a bit different approach, cause we would be basically using different tree, so we keep this note in just as a “hack”.

2ª The root is black.

This rule might seem like a very important one, but overall is not. You can safely omit this rule, but you also need to deal with the consequences.

Let's refresh our memory with the algorithm of insert fixup:

WHILE z.p.color == Red
  IF z.p == z.p.p.left
    y = z.p.p.right

    IF y.color == Red
      z.p.color = Black
      y.color = Black
      z.p.p.color = Red
      z = z.p.p
    ELSE
      IF z == z.p.right
        z = z.p
        Left-Rotate(T, z)
      z.p.color = Black
      z.p.p.color = Red
      Right-Rotate(T, z.p.p)
  ELSE (same as above with “right” and “left” exchanged)

T.root.color = Black

:::tip

If you have tried to implement any of the more complex data structures, such as red-black trees, etc., in a statically typed language that also checks you for NULL-correctness (e.g. mypy or even C# with nullable reference types), you might have run into numerous issues in the cases where you are 100% sure that you cannot obtain NULL because of the invariants, but the static type checking doesn't know that.

The issue we hit with the insert fixup is very similar.

:::

You might not realize the issue at the first sight, but the algorithm described with the pseudocode above expects that the root of the red-black tree is black by both relying on the invariant in the algorithm and afterwards by enforcing the black root property.

If we decide to omit this condition, we need to address it in the pseudocodes accordingly.

Usual algorithm with black root	Allowing red root

3ª Every leaf (nil) is black.

Now, this rule is a funny one. What does this imply and can I interpret this in some other way? Let's go through some of the possible ways I can look at this and how would they affect the other rules and balancing.

We will experiment with the following tree:

We should start by counting the black nodes from root to the nil leaves based on the rules. We have multiple similar paths, so we will pick only the interesting ones.

1. What happens if we do not count the nil leaves?
2. What happens if we consider leaves the nodes with no descendants, i.e. both of node's children are nil?
3. What happens if we do not count the nil leaves, but consider nodes with at least one nil descendant as leaves?

path	black nodes	1ª idea	2ª idea	3ª idea
3 → 1 → 0 → nil	4	3	4	3
3 → 5 → 7 → 8 → nil	4	3	-	3
3 → 5 → 7 → 8 → 9 → nil	4	3	4	3

First idea is very easy to execute and it is also very easy to argue about its correctness. It is correct, because we just subtract one from each of the paths. This affects all paths and therefore results in global decrease by one.

Second idea is a bit more complicated. We count the nils, so the count is 4 as it should be. However, there is one difference. Second path no longer satisfies the condition of a leaf. Technically it relaxes the 5th rule, because we leave out some of the nodes. We should probably avoid that.

:::caution

With the second idea, you may also feel that we are “bending” the rules a bit, especially the definition of the “leaf” nodes.

Given the definition of the red-black tree, where nil is considered to be an external node, we have decided that bending it a bit just to stir a thought about it won't hurt anybody. 😉

:::

4ª If a node is red, then both its children are black.

This rule might seem rather silly on the first look, but there are 2 important functions:

1. it allows the algorithms to “notice” that something went wrong (i.e. the tree needs to be rebalanced), and
2. it holds the balancing and height of the tree “in check” (with the help of the 5th rule).

When we have a look at the algorithms that are used for fixing up the red-black tree after an insertion or deletion, we will notice that all the algorithms need is the color of the node.

How come it is the only thing that we need? How come such naïve thing can be enough?

Let's say we perform an insertion into the tree… We go with the usual and pretty primitive insertion into the binary-search tree and then, if needed, we “fix up” broken invariants. How can that be enough? With each insertion and deletion we maintain the invariants, therefore if we break them with one operation, there's only one path on which the invariants were felled. If we know that rest of the tree is correct, it allows us to fix the issues just by propagating it to the root and abusing the siblings (which are, of course, correct red-black subtrees) to fix or at least partially mitigate the issues and propagate them further.

Let's assume that we do not enforce this rule, you can see how it breaks the balancing of the tree below.

Enforcing this rule	Omitting this rule

We can create a big subtree with only red nodes and even when keeping the rest of the rules maintained, it will break the time complexity. It stops us from “hacking” the black height requirement laid by the 5th rule.

5ª For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.

As it was mentioned, with the 4th rule they hold the balancing of the red-black tree.

:::tip

An important observation here is the fact that the red-black tree is a height-balanced tree.

:::

Enforcing this rule (together with the 4th rule) keeps the tree balanced:

1. 4th rule makes sure we can't “hack” this requirement.
2. This rule ensures that we have “similar”² length to each of the leaves.

:::tip AVL tree

You might have heard about an AVL tree before. It is the first self-balanced tree to be ever introduced and works in a very similar nature as the red-black tree, the only difference is that it does not deal with the black height, but the height in general.

If you were to compare AVL with the red-black tree, you can say that AVL is much more strict while red-black tree can still maintain the same asymptotic time complexity for the operations, but having more relaxed rules.

:::

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1. CORMEN, Thomas. Introduction to algorithms. Cambridge, Mass: MIT Press, 2009. isbn 9780262033848. ↩︎

2. red nodes still exist ↩︎