树上的尾递归 [英] Tail-recursion on trees

问题描述

我有一个数据结构，

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

我想写一个函数，以某种顺序遍历这棵树.它做什么并不重要，所以它可能是一个 treefold : ('a * 'b -> 'b) ->'b ->'一棵树 ->'b.我可以这样写这个函数:

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

但是因为在最后一种情况下我不可避免地有两个分支，所以这不是尾递归函数.

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

是否有可能创建一个，即允许扩展类型签名，并且成本是多少?我也想知道它是否值得尝试；也就是说，它在实践中是否会带来任何速度优势?

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?

推荐答案

您可以使用 continuation-passing 样式实现尾递归树状折叠:

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)

例如:

fun sumtree t = treefold op+ t 0

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

val n = sumtree t1

按预期结果 n = 6.

results in n = 6 as expected.

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