在f#中对多路树进行折叠/递归 [英] Fold / Recursion over Multiway Tree in f#

问题描述

我正在尝试将Brian's Fold应用于二叉树( http: //lorgonblog.wordpress.com/2008/04/06/catamorphisms-part-two/)来申请Multiway树.

I am trying to adapt Brian's Fold for Bianary Trees (http://lorgonblog.wordpress.com/2008/04/06/catamorphisms-part-two/) to apply for Multiway trees.

摘自Brian的博客:

Summarizing from Brian's Blog:

数据结构:

type Tree<'a> =  
    | Node of (*data*)'a * (*left*)Tree<'a> * (*right*)Tree<'a>  
    | Leaf 

let tree7 = Node(4, Node(2, Node(1, Leaf, Leaf), Node(3, Leaf, Leaf)),  
                    Node(6, Node(5, Leaf, Leaf), Node(7, Leaf, Leaf)))

二叉树折叠功能

let FoldTree nodeF leafV tree =   
    let rec Loop t cont =   
        match t with   
        | Node(x,left,right) -> Loop left  (fun lacc ->    
                                Loop right (fun racc ->   
                                cont (nodeF x lacc racc)))   
        | Leaf -> cont leafV   
    Loop tree (fun x -> x) 

示例

let SumNodes = FoldTree (fun x l r -> x + l + r) 0 tree7
let Tree6to0 = FoldTree (fun x l r -> Node((if x=6 then 0 else x), l, r)) Leaf tree7

多路树版本 [无法(完全)正常运行:

数据结构

type MultiTree = | MNode of int * list<MultiTree>

let Mtree7 = MNode(4, [MNode(2, [MNode(1,[]); MNode(3, [])]);  
                    MNode(6, [MNode(5, []); MNode(7, [])])])

折叠功能

let MFoldTree nodeF leafV tree = 
    let rec Loop  tree cont =   
        match tree with   
        | MNode(x,sub)::tail -> Loop (sub@tail) (fun acc -> cont(nodeF x acc))
        | [] -> cont leafV
    Loop  [tree] (fun x -> x) 

示例1 返回28-似乎有用

Example 1 Returns 28 - seems to work

let MSumNodes = MFoldTree (fun x acc -> x + acc) 0 Mtree7

示例2

不运行

let MTree6to0 = MFoldTree (fun x acc -> MNode((if x=6 then 0 else x), [acc])) Mtree7

起初我以为MFoldTree在某个地方需要一个map.something，但是我改为使用@运算符.

Initially I thought the MFoldTree needed a map.something somewhere but I got it to work with the @ operator instead.

在第二个示例上提供任何帮助，或者更正我在MFoldTree函数中所做的任何事情，都是很棒的！

Any help on the second example and or correcting what I've done in the MFoldTree function would be great!

欢呼

dusiod

推荐答案

另一个解决方案可能是

let rec mfold f a (MNode(x,s)) = f (List.fold (fun a t -> mfold f a t) a s) x

真的，我们可以将树视为线性结构(将其折叠).

really, we can treat tree as a lineal struct (to fold it).

用例

> mfold (+) 0 Mtree7;;
val it : int = 28

过滤器与常规折叠相同(因为mfold是常规折叠):

Filter is the same with normal fold (because mfold is a normal fold):

> mfold (fun a x -> if x = 6 then a else x + a) 0 Mtree7;;
val it : int = 22

该功能可能是通用的(因为List.fold，Array.fold，...可能是通用的).

That function could be generic (as List.fold, Array.fold, ... could be generics).

，但是第二个意图是返回修改后的整个树，以使任何具有值6的节点现在具有值0"

但这不是fold计算，是map！

您可以轻松进行(再次处理，作为线性结构)

You can do easilly (treating, again, as a lineal struct)

let rec mmap f (MNode(x,s)) = MNode(f x, List.map (mmap f) s)

用例

> mmap (fun x -> if x=6 then 0 else x) Mtree7;;
val it : MultiTree =
  MNode
    (4,
     [MNode (2,[MNode (1,[]); MNode (3,[])]);
      MNode (0,[MNode (5,[]); MNode (7,[])])])

同样，我建议对每个可能的列表容器(Seq，List，Array，...)执行此操作，它使用户可以在上下文中选择最佳策略.

Again, I suggest to do it for each possible list container (Seq, List, Array, ...), it enable to user select best strategy in context.

注意:

• 我是F#的新手，如果有些错误，请原谅.
• 堆栈大小应该没问题，堆栈级别等于树的深度.

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