CODEWITHSUNDEEP
functional-programmingsml
sshine
sshine

Reputation: 16105

Tail-recursion on trees

I have a data structure,

datatype 'a tree = Leaf | Branch of 'a tree * 'a * 'a tree

and I want to write a function that traverses this tree in some order. It doesn't matter what it does, so it could be a treefold : ('a * 'b -> 'b) -> 'b -> 'a tree -> 'b. I can write this function like this:

fun treefold f acc1 Leaf = acc1
  | treefold f acc1 (Branch (left, a, right)) =
    let val acc2 = treefold f acc1 left
        val acc3 = f (a, acc2)
        val acc4 = treefold f acc3 right
    in acc4 end

But because I inevitably have two branches in the last case, this is not a tail-recursive function.

Is it possible to create one that is, given the type signature is allowed to be expanded, and at what cost? I also wonder if it's even worth trying; that is, does it give any speed benefits in practice?

Upvotes: 4

Views: 1622

Answers (2)

kfl
kfl

Reputation: 56

Like @seanmcl writes, the systematic way to convert a function to be tail-recursive is to use continuation-passing style.

After that you probably want to reify your continuations and use a more concrete data type, like a list for instance:

fun treefoldL f init tree =
    let fun loop Leaf acc [] = acc
          | loop Leaf acc ((x, right) :: stack) =
            loop right (f(x,acc)) stack
          | loop (Branch (left, x, right)) acc stack =
            loop left acc ((x, right) :: stack)
    in  loop tree init [] end

Upvotes: 2

seanmcl
seanmcl

Reputation: 9946

You can achieve a tail-recursive treefold using continuation-passing style:

fun treefold1 f Leaf acc k = k acc
  | treefold1 f (Branch (left, a, right)) acc k =
    treefold1 f left acc (fn x => treefold1 f right (f(a, x)) k)

fun treefold f t b = treefold1 f t b (fn x => x)

For example:

fun sumtree t = treefold op+ t 0

val t1 = Branch (Branch(Leaf, 1, Leaf), 2, Branch (Leaf, 3, Leaf))

val n = sumtree t1

results in n = 6 as expected.

Upvotes: 5

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