So, we've built up some

pretty nifty binary trees - we can use the binary tree both as the basis of an

implementation of a set, or as an implementation of a dictionary. But our

implementation has had one major problem: it's got absolutely no way to

maintain balance. What that means is that depending on the order in which

things are inserted to the tree, we might have excellent performance, or we

might be no better than a linear list. For example, look at these trees. As

you can see, a tree with the same values can wind up quite different. In a

good insert order, you can wind up with a nicely balanced tree: the minimum

distance from root to leaf is 3; the maximum is 4. On the other hand, take the

same values, and insert them in a different order and you get a rotten tree;

the minimum distance from root to leaf is 1, and the maximum is 7. So

depending on luck, you can get a tree that gives you good performance, or one

that ends up giving you no better than a plain old list. Playing with a bit of

randomization can often give you reasonably good performance on average - but if

you're using a tree, it's probably because O(n) complexity is just too high. You

want the O(lg n) complexity that you'll get from a binary tree - and

not just sometimes.

To fix that, you need to change the structure a bit, so that as you insert

things, the tree stays balanced. There are several different approaches to how

you can do this. The one that we're going to look at is based on labeling nodes

in ways that allow you to very easily detect when a serious imbalance is

developing, and then re-arrange the tree to re-balance it. There are two

major version of this, called the AVL tree, and the red-black tree. We're going

to look at the red-black. Building a red-black tree is as much a

lesson in data structures as it is in Haskell, but along with learning about

the structure, we'll see a lot about how to write code in Haskell, and particularly

about how to use pattern-matching for complex structures.

We'll start with a basic definition. A red-black tree is a normal binary

search tree, except that each node is assigned a *color*, which is

either red or black, and there are several invariant properties that must

hold about the coloring of the tree:

- The root of the tree is always black.
- All branches of a tree end in a null which is
*black*. - All children of red nodes are black.
- For all nodes in the tree, all downward paths from the node to a leaf contain the

same number of*black*nodes.

If these invariants are maintained, they guarantee that tree is *almost* balanced:

for the entire tree, and every subtree of it, the *longest* path from the root to

a leaf is no more than twice the *shortest* path from the root to a leaf.

We'll start by writing the Haskell type declaration for a red/black tree. We'll just do it

as a tree of ordered values to keep things simple.

>data Color = Red | Black deriving (Eq, Show) > >data (Ord a) => RedBlackTree a = RBTNode a Color (RedBlackTree a) (RedBlackTree a) > | RBTEmpty deriving (Eq,Show)

Also, for convenience, we'll write a couple of accessor functions that we'll

use later on. Something interesting to note about these accessors is that they use

*non-exhaustive patterns*: there are values of type `RedBlackTree a`

for which these functions are undefined. If you call any of these accessors on

a tree whose value is `RBTEmpty`

, you'll get a runtime error.

It is, at the very least, considered bad style to write non-exhaustive

functions. It's actually a way of cheating the type system. You're claiming

that you're writing a function from type T to type U, but in fact, there are

values of T for which the function won't work: the real type of the function

is T' -> U, where T' is a subset of T. But you can't say that in Haskell - so

you're cheating. To be more concrete, you're writing functions like

`rbtLeftChild`

which *claims* that for any red-black tree

passed to it, it will return a valid red-black tree. But in fact, that's only

true for the subset of red-black trees that were built with the `RBTNode`

constructor; for other values, the function will fail.

The best solution to make it exhaustive would be to use the `Maybe`

type to allow you to return a valid value for all trees passed as inputs. But that

would make the code *much* more complex, unless we used monads - and we're

not ready for monads yet.

> >rbtLeftChild :: (Ord a) => RedBlackTree a -> RedBlackTree a >rbtLeftChild (RBTNode _ _ l _) = l > >rbtRightChild :: (Ord a) => RedBlackTree a -> RedBlackTree a >rbtRightChild (RBTNode _ _ _ r) = r > >rbtValue :: (Ord a) => RedBlackTree a -> a >rbtValue (RBTNode v _ _ _) = v > >rbtColor :: (Ord a) => RedBlackTree a -> Color >rbtColor (RBTNode _ c _ _) = c >rbtColor RBTEmpty = Black

Inserting data into the tree is where things get interesting. It starts

off the same as how you insert into a typical BST: search for the correct

position, and then insert the value as a new leaf node. But in a red-black tree,

the new node needs a color. New nodes are always red - so you're inserting a red

node. Now you need to check to make sure that you're not violating any of the

tree invariants. If you are, then you need to fix it.

To

keep things reasonably clean and separate, we'll use the tail-calling version

of tree insert, and then tail-call a rebalance function when the basic insert is

complete. Rebalance will fix the balance of the tree, and do the tree

re-assembly as it climbs up the tree.

>rbtInsert :: (Ord a) => RedBlackTree a -> a -> RedBlackTree a >rbtRebalance :: (Ord a) => RedBlackTree a -> [RedBlackTree a] -> RedBlackTree a >--rbtRebalance focus ancestors > >rbtInsert node v = > rbtInsertTailCall node v [] > >rbtInsertTailCall node@(RBTNode v color left right) newval path > | v > newval = rbtInsertTailCall left newval (node:path) > | otherwise = rbtInsertTailCall right newval (node:path) >rbtInsertTailCall RBTEmpty v path = > rbtRebalance (RBTNode v Red RBTEmpty RBTEmpty) path

All over the place as we rebalance the tree, we'll have places where we want to

"rebuild" nodes to patch in the insertion change; as usual, we separate that into

its own function.

>-- Reconstruct takes a child node and a parent node, and creates a replacement >-- for the parent node with the child in the appropriate position. It allows >-- the color of the new node to be specified. >reconstructNode node@(RBTNode v c l r) parent@(RBTNode pv pc pl pr) color = > if (pv > v) > then (RBTNode pv color node pr) > else (RBTNode pv color pl node)

Now, we need to think about what we're going to do to keep the tree balanced as we walk

back up the insertion path fixing the tree. There are two things we can do to make the

tree respect the invariants: we can re-color nodes, or we can *pivot* subtrees.

Pivoting a tree is an interesting operation - it's a process of swapping a node and one of

its children to rotate a section of the tree. Suppose we have a binary search tree like the one

in the diagram to the side. It's poorly balanced; it's got only one node to its left, but 7

nodes to its right. To correct this by pivoting, what we'll do is take node 6 - currently a

child of the root, and rotate the tree counterclockwise around it, so that 6 becomes the root,

the old root (2) becomes the left child of 6, and the old left child of 6 (node 4) becomes the

*right* child of the old root.

So after the pivot, our tree looks like this. This

operation was a *left* pivot; a right pivot does the same kind of thing, but rotating

the tree clockwise instead of counterclockwise.

So let's go ahead and write the pivot operations. We'll write two pivot

functions: one for each direction. We'll pass the pivot operation

a subtree whose root and child in the appropriate direction are to be rotated. In addition,

we'll also add a parameter for managing the color of the new root node. In some cases,

we'll want to swap the colors of the nodes being moved; in other cases, we won't. So we'll

put a boolean parameter in to specify whether or not to swap the colors.

> -- pivot left tree at root; second parent indicates whether or not to swap > -- colors of the nodes that are being moved. >rbtPivotLeft :: (Ord a) => RedBlackTree a -> Bool -> RedBlackTree a >rbtPivotLeft (RBTNode rootval rootcolor sib (RBTNode focval foccolor focleft focright)) swap = > (RBTNode focval newrootcolor oldroot focright) where > newrootcolor = if swap then rootcolor else foccolor > oldrootcolor = if swap then foccolor else rootcolor > oldroot = RBTNode rootval oldrootcolor sib focleft > > >rbtPivotRight (RBTNode rootval rootcolor (RBTNode focval foccolor focleft focright) sib) swap = > (RBTNode focval newrootcolor focleft oldroot) where > newrootcolor = if swap then rootcolor else foccolor > oldrootcolor = if swap then foccolor else rootcolor > oldroot = RBTNode rootval oldrootcolor focright sib >

So, let's try taking a look at how the pivots work. First, we need to construct

some trees to rebalance. We'll just do it manually, since the insert code isn't properly

finished yet.

>twentyseven = RBTNode 27 Black RBTEmpty RBTEmpty >twentytwo = RBTNode 22 Black RBTEmpty RBTEmpty >twentyfive = RBTNode 25 Black twentytwo twentyseven >sixteen = RBTNode 16 Black RBTEmpty RBTEmpty >twenty = RBTNode 20 Black sixteen twentyfive >twelve = RBTNode 12 Black RBTEmpty RBTEmpty >fifteen = RBTNode 15 Black twelve twenty >two = RBTNode 2 Black RBTEmpty RBTEmpty >seven = RBTNode 7 Black RBTEmpty RBTEmpty >five = RBTNode 5 Black two seven >ten = RBTNode 10 Black five fifteen

That produces a unbalanced binary tree that looks like this:

RBTNode 10 Black (RBTNode 5 Black -- 10left (RBTNode 2 Black RBTEmpty RBTEmpty) -- 5 left (RBTNode 7 Black RBTEmpty RBTEmpty)) -- 5 right (RBTNode 15 Black -- 10 right (RBTNode 12 Black RBTEmpty RBTEmpty) -- 15 left (RBTNode 20 Black -- 15 right (RBTNode 16 Black RBTEmpty RBTEmpty) -- 20 left (RBTNode 25 Black -- 20 right (RBTNode 22 Black RBTEmpty RBTEmpty) -- 25 left (RBTNode 27 Black RBTEmpty RBTEmpty)))) -- 25 right

Let's do a quick test, and try doing a left pivot on the root.

*Main> rbtPivotLeft ten False RBTNode 15 Black (RBTNode 10 Black (RBTNode 5 Black (RBTNode 2 Black RBTEmpty RBTEmpty) (RBTNode 7 Black RBTEmpty RBTEmpty)) (RBTNode 12 Black RBTEmpty RBTEmpty)) (RBTNode 20 Black (RBTNode 16 Black RBTEmpty RBTEmpty) (RBTNode 25 Black (RBTNode 22 Black RBTEmpty RBTEmpty) (RBTNode 27 Black RBTEmpty RBTEmpty))) *Main>

Cleaned up, that looks like this:

RBTNode 15 Black (RBTNode 10 Black (RBTNode 5 Black (RBTNode 2 Black RBTEmpty RBTEmpty) (RBTNode 7 Black RBTEmpty RBTEmpty)) (RBTNode 12 Black RBTEmpty RBTEmpty)) (RBTNode 20 Black (RBTNode 16 Black RBTEmpty RBTEmpty) (RBTNode 25 Black (RBTNode 22 Black RBTEmpty RBTEmpty) (RBTNode 27 Black RBTEmpty RBTEmpty)))

Much better - that's much closer to a balanced tree! So now that we know how to do the

pivot, and we've seen that it works correctly, we can look at building the rebalance code.

With pivots out of the way, we can start looking at how to decide what

operations to do to rebalance the tree. When we're doing an insert, we end up

inserting a red node on the bottom of the tree. It's got two children, both

null, which are considered black. If the parent of our new node is black, then

everything is fine; we haven't altered the number of black nodes on any path

from a node to a leaf. So we're done. But if the parent is red, then we've got

a red child of a red node, so we need to do some fixing.

Fixing an imbalance in a red-black tree can (and in fact often will)

trigger a cascade of changes. But part of what makes the structure

so elegant is that we only need to look at the *local* structure

immediately around the new insert; and then when we've corrected that,

there's only *one* place where the next problem could be. In every case

where we're rebalancing, we can look at a specific problem, and fix it, and

then immediately move to where the next potential problem is. To code

this, we'll look at in terms of a *focal node*, which is

the node causing the immediate problem we're fixing; and we'll fix the problem

by looking at the local context of the focus.

The potential cases we can encounter are:

- The focal node is the root of the tree. In that case, we make it

black. That adds one black node to every path in the tree, which

leaves us with a valid tree, so we're done. - The focal node is red, but has a black parent. Again, that's fine. No

problem. - The focal node is red; it's parent is also red. Then we need to look at

its*uncle*; that is, the node that is the sibling of its parent. If

both the new node, the parent and the uncle are all red, then we change the

color of the parent and uncle to black, and the grandparent to red. After

this, the grandparent becomes the focal node, and we continue to do our

tree-fixing with the new focus. - Here's where it gets a bit messy. If the focal node and its parent are both red,

but the uncle is black, then we're going to need to pivot. Getting the pivot right

is tricky. There are four cases:- The focal node is the
*right*child of its parent, and the

parent is the*left*node of the grandparent, then we do a

*left*pivot of the focal node and its parent, and the former

parent becomes the new focal node. - The focal node is the
*left*child of its parent, and the

parent is the*right*child of the grandparent, then we do a

*right*pivot of the focal node and its parent, and the former

parent becomes the new focus. - The focal node is the left child of its parent, and the parent is

the left child of the grandparent. Then we do a*right*pivot

of the parent and the grandparent and swap the colors of the parent

and grandparent. The parent becomes the focus. - The focal node is the right child of its parent, and the parent

is the right child of the grandparent. Then we do a*left*

pivot of the parent and the grandparent and swap the colors of the

parent and grandparent. The parent becomes the focus.

- The focal node is the

Ok, there's the algorithm for rebalancing. How can we code it in Haskell?

We've got a list of the nodes from the insertion path, in leaf to root order.

When we look at the rebalance, we can see that there are a bunch of different

cases which we can separate via pattern matching:

- The focus is the root of the tree. We can select this case by

using an empty list for the pattern for the ancestors parameter. Once

we've gotten to the root, the tree is balanced, and the only corrective

thing we may need to do is make the root black. So:>-- Root is focus; no matter what color it is, just make it black >rbtRebalance (RBTNode v _ left right) [] = RBTNode v Black left right >rbtRebalance node@(RBTNode v _ left right) (parent@(RBTNode pv pc pl pr):[]) > | pv > v = RBTNode pv pc node pr > | otherwise = RBTNode pv pc pl node

- Also very simple is the case where the focus is black. In that case,

we don't need to do anything except patch in the insert, and continue up

the tree. Again, we can select that case just by pattern matching.>-- black node - just patch in the change, and climb. > >rbtRebalance focus@(RBTNode fv Black left right) (parent@(RBTNode pv pc pl pr):ancestors) > | pv > fv = rbtRebalance (RBTNode pv pc focus pr) ancestors > | otherwise = rbtRebalance (RBTNode pv pc pl focus) ancestors >

- Next, we've got the case of a red node with a black parent. We can

identify it by using "`RBTNode v Red left right`

" as a pattern for the

focus, and "`RBTNode _ Black _ _`

" as a pattern for the parent. A

red node with a black parent is OK, as long as the subtree under the red is

balanced; and since we're balancing from the bottom up, we know that

everything beneath this node is balanced. So:>rbtRebalance focus@(RBTNode fv Red left right) (parent@(RBTNode pv Black pl pr):ancestors) = > rbtRebalance (reconstructNode focus parent Black) ancestors

- Now we're getting to the interesting cases, which are the cases where

both the node and its parent are red. We can separate two cases here: cases

where we'll fix using a pivot, and cases where we'll fix using a

recoloring. The way to distinguish them is by looking at the*uncle*

of the focus node; that is, the sibling of the nodes parent. The red-red

case is complicated enough that instead of writing out huge pattern

expressions, we'll simplify it by separating the function into several

layers of calls, each of which does a phase of the pattern match. We want

to separate out the cases where we've got a red node with a red parent and

a red uncle, and the cases where we've got a red node with a red parent and

a black uncle.If the focus, its parent, and its uncle are all red, then we're in a

recoloring case; if the focus and its parent are red, and the uncle is black,

then we're in a pivot case.>rbtRebalance focus@(RBTNode v Red left right) (parent@(RBTNode _ Red _ _):ancestors) = > rebalanceRedRedNode focus parent ancestors

To be able to recognize sub-cases when we have a red node/red parent, we need

to be able to look at the path from the grandparent to the focus, and the color of the uncle. So

we'll write some helper functions to get those.>uncleColor node parent grandparent = > if (parent == rbtLeftChild grandparent) > then rbtColor (rbtRightChild grandparent) > else rbtColor (rbtLeftChild grandparent) > >data TwoStepPath = LeftLeft | LeftRight | RightLeft | RightRight > >pathFromGrandparent :: (Ord a) => RedBlackTree a -> RedBlackTree a -> RedBlackTree a -> TwoStepPath >pathFromGrandparent node@(RBTNode v _ l r) parent@(RBTNode pv _ pl pr) grand@(RBTNode gv _ gl gr) > | pv < gv && v < pv = LeftLeft > | pv >= gv && v < pv = RightLeft > | pv < gv && v >= pv = LeftRight > | pv >= gv && v >= pv = RightRight

To actually handle the red node/red parent, first we separate out the

case where the red parent is the root of the tree - there are no more ancestors

on the insertion path. In that case, we can just climb to root, and do the correction

from there.> >-- node is red, parent is red, but parent is root: just go to parent(root), and fix >-- from there. >rebalanceRedRedNode focus@(RBTNode fv fc fl fr) parent@(RBTNode pv pc pl pr) [] = > rbtRebalance (reconstructNode focus parent Red) []

Otherwise, we need to check whether the uncle was red or black. If it was black,

we do a recolor correction; if it was red, we figure out what kind of pivot to do. We'll

use a bunch of helper functions to make it easy.>rebalanceRedRedNode focus parent (grand@(RBTNode gv gc gl gr):ancestors) = > if (uncleColor focus parent grand) == Red > then recolorAndContinue focus parent grand ancestors > else case (pathFromGrandparent focus parent grand) of > LeftLeft -> rbtRebalance (pivotGrandparentRight focus parent grand) ancestors > LeftRight -> rbtRebalance (pivotParentLeft focus parent) (grand:ancestors) > RightLeft -> rbtRebalance (pivotParentRight focus parent) (grand:ancestors) > RightRight -> rbtRebalance (pivotGrandparentLeft focus parent grand) ancestors

The code above is really just using patterns for case selection. The

actual work is in the helper functions that get called. They're all simple

functions. First, we have some custom pivot functions - one for each direction

for pivoting around a parent (the cases where the node is left of the parent,

and the parent is right of the grandparent, or vise versa), and one for each

direction pivoting around a grandparent (both node and parent are left

children, or both are right children).>pivotGrandparentLeft node parent@(RBTNode pv pc pl pr) grand@(RBTNode gv gc gl gr) = > rbtPivotLeft (RBTNode gv gc gl (RBTNode pv pc pl node)) True > >pivotGrandparentRight node parent@(RBTNode pv pc pl pr) grand@(RBTNode gv gc gl gr) = > rbtPivotRight (RBTNode gv gc (RBTNode pv pc node pr) gr) True > >pivotParentLeft node parent@(RBTNode pv pc pl pr) = > rbtPivotLeft (RBTNode pv pc pl node) False > >pivotParentRight node parent@(RBTNode pv pc pl pr) = > rbtPivotRight (RBTNode pv pc node pr) False

And a function to do the recoloring for when the uncle was red:

>recolorAndContinue focus@(RBTNode v c l r) parent@(RBTNode pv pc pl pr) grand@(RBTNode gv gc gl gr) ancestors = > let path = pathFromGrandparent focus parent grand > uncle = (case path of > LeftLeft -> gr > LeftRight -> gr > RightLeft -> gl > RightRight -> gl) > newUncle = if (uncle == RBTEmpty) > then RBTEmpty > else (RBTNode (rbtValue uncle) Black (rbtLeftChild uncle) (rbtRightChild uncle)) > newparent = reconstructNode focus parent Black > newGrandParent = (case path of > LeftLeft -> (RBTNode gv Red newparent newUncle) > LeftRight -> (RBTNode gv Red newparent newUncle) > RightLeft -> (RBTNode gv Red newUncle newparent) > RightRight -> (RBTNode gv Red newUncle newparent)) > in rbtRebalance newGrandParent ancestors

And that, finally, is it. For the binary search tree without balancing code, the

worst case is inserting a list of values in order. So let's try that, to see how

well this works.

*Main> foldl (\ x y -> rbtInsert x y) RBTEmpty [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16] RBTNode 4 Black (RBTNode 2 Black (RBTNode 1 Black RBTEmpty RBTEmpty) (RBTNode 3 Black RBTEmpty RBTEmpty)) (RBTNode 8 Red (RBTNode 6 Black (RBTNode 5 Black RBTEmpty RBTEmpty) (RBTNode 7 Black RBTEmpty RBTEmpty)) (RBTNode 12 Black (RBTNode 10 Red (RBTNode 9 Black RBTEmpty RBTEmpty) (RBTNode 11 Black RBTEmpty RBTEmpty)) (RBTNode 14 Red (RBTNode 13 Black RBTEmpty RBTEmpty) (RBTNode 15 Black RBTEmpty (RBTNode 16 Red RBTEmpty RBTEmpty)))))

Since that's completely illegible, let's clean it up, and look at it in picture form:

The shortest path from root to leaf is [4,2,1]; the longest is [4,8,12,14,15,16]. Just

like we promised: the longest is no more than twice the shortest. It's a pretty good

search tree, and the rebalancing work isn't terribly expensive, and amortizes nicely

over a long run of inserts. The insert time ends up amortizing to **O**(lg n), just like the simple

binary search tree insert.

- Log in to post comments

OT, but tip for bad math: http://kstp.com/news/stories/S1285386.shtml

Mark, are you familiar with this paper?

http://www.eecs.usma.edu/webs/people/okasaki/jfp99.ps

The gist of it is that imperative implementations of RB trees tend to be more complicated because destructive updates to the tree are possible. In functional languages, you can't do that, so red-black might as well be boiled down to a very simple pattern match of four cases. Not totally sure what your sources are but it looks closer to a reproduction of an imperative algorithm.

There is a good example of using GADT's to certify the correctness of some red-black tree operations:

http://www.reddit.com/r/programming/comments/w1oz/how_are_gadts_useful_…

I think the example may originate from a paper someplace but I don't know which one.

I like your diagrams - what are you using to draw the trees? I've been coding binary trees in Haskell too, and I'm looking for a way to visualize them.

The best I can do so far is dump a representation of the tree to a text file in DOT format and run graphviz on it.

Mark woud you mind making a post about zippers or tries ? It seems that those are the most used data structures.

I remember barely understanding these in university. I hope i never need to understand them again.

Hmm. This is reminiscent of how I teach my middle school students "how to factor." Draw two factorizations of, say, 24 on the board, and point out how the "balanced" one is faster than the "unbalanced" one. Maybe, someday, one of them will be a computer programmer and this will resonate! ;o/

Dan McKinley makes an excellent point. From p. 27 of Chris Okasaki's Purely Functional Data Structures, the book that expands upon the work in his thesis:

The book is one I think that any Haskell programmer who builds his own data structures should own.