Ocaml列表到树

Question

我想编写一个函数load: 'a option list -> 'a tree,它从给定列表中恢复二进制树,其中包含后缀顺序的元素.如果列表不代表任何树,则您的函数应该引发异常Load(您必须先声明它).

树被定义为:

type ‘a btree = L of ‘a | N of ‘a btree * ‘a btree

到目前为止,我所做的是:

exception Load ;;

let rec load l =
    match l with
        | [] -> raise Load
        | Some x::_ -> L x
        | None::t -> N(?,load t)
;; 

以下是输入列表的示例:

[Some 2; Some 0; None; Some 5; Some 9; Some 20; None; None; None]

如何做到这一点真是太棘手了.由于列表是后缀顺序,我想知道是否更好地使用List.foldright函数??

Answer 1

我认为你应该更加努力地完成你的作业(当然有一些有趣的东西需要学习),但是你显然不在乎:

let load input =
  let rec load stack = function
    | [] -> stack
    | Some elem :: rest -> load (L elem :: stack) rest
    | None :: rest ->
      match stack with
        | right :: left :: stack -> load (N(left, right) :: stack) rest
        | [] | [_] -> failwith "incorrect node arity"
  in
  match load [] input with
    | res :: [] -> res
    | [] -> failwith "no input"
    | _ :: _ :: _ -> failwith "remaining nodes"