---
id: rules
title: On the rules of the red-black tree
description: |
  Shower thoughts on the rules of the red-black tree.
tags:
- red-black trees
- balanced trees
last_update:
  date: 2023-06-10
---

import ThemedSVG from "@site/src/components/ThemedSVG";

## 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 rules[^1]:
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:

<ThemedSVG
  source="/files/ib002/rb-trees/rules/rb_height"
  alt="Red-black tree with black height"
/>

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 |
| :-----------------------------: | :---------------: |
| <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/br_0" alt="1ª insertion" /> | <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/rr_0" alt="1ª insertion" /> |
| <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/br_1" alt="2ª insertion" /> | <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/rr_1" alt="2ª insertion" /> |
| <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/br_2" alt="3ª insertion" /> | <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/rr_2" alt="3ª insertion" /> |
| <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/br_3" alt="4ª insertion" /> | <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/rr_3" alt="4ª insertion" /> |
| <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/br_4" alt="5ª insertion" /> | <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/rr_4" alt="5ª insertion" /> |
| <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/br_5" alt="6ª insertion" /> | <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/rr_5" alt="6ª insertion" /> |
| <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/br_6" alt="7ª insertion" /> | <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/rr_6" alt="7ª insertion" /> |
| <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/br_7" alt="8ª insertion" /> | <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/rr_7" alt="8ª insertion" /> |
| <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/br_8" alt="9ª insertion" /> | <ThemedSVG source="/files/ib002/rb-trees/rules/red-root/rr_8" alt="9ª insertion" /> |

## 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:
<ThemedSVG source="/files/ib002/rb-trees/rules/rb" />

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 `nil`s, 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. :wink:

:::

## 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.

import Tabs from '@theme/Tabs';
import TabItem from '@theme/TabItem';

<Tabs>
  <TabItem value="enforcing" label="Enforcing this rule">
    <ThemedSVG source="/files/ib002/rb-trees/rules/red-node-black-children/correct" />
  </TabItem>

  <TabItem value="omitting" label="Omitting this rule">
    <ThemedSVG source="/files/ib002/rb-trees/rules/red-node-black-children/incorrect" />
  </TabItem>
</Tabs>

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.

:::

[^1]: CORMEN, Thomas. Introduction to algorithms. Cambridge, Mass: MIT Press, 2009. isbn 9780262033848.
[^2]: red nodes still exist