Question

如何将树转换为列表：到目前为止，我有以下代码，但它提出了几个问题：

type 'a tree = Lf | Br of 'a * 'a tree * 'a tree;;

let rec elementRight xs = function
  | LF ->false
  | Br(m,t1,t2) -> if elementRight xs t1 = false then t1::xs else element xs t1;; //cannot find element
let rec elementLeft xs = function
  | LF ->false
  | Br(m,t1,t2) -> if elementLeft xs t2 = false then t2::xs else element xs t2 ;; //cannot find element
let rec element xs = function
  | LF ->[]
  | Br(m,t1,t2) -> xs::(elementRight xs t1)::(elementRight xs t2)::(elementLeft xs t1)::(elementLeft xs t2);;

Answer 1

您的代码存在许多问题：

你不应该在行的末尾有;;（我猜这意味着你要复制并粘贴到repl中，你应该使用fsx文件代替并使用“发送”复制“）。
这：| LF ->false正在返回一个bool，而这个：| Br(m,t1,t2) -> if elementRight xs t1 = false then t1::xs else element xs t1正在返回'a list。表达式只能有一个类型，因此返回两个是编译错误。我猜你真正要做的是让叶子返回[]，然后在你的分支案例中检查空列表是这样的：

let rec elementRight xs = function
  | LF ->[]
  | Br(m,t1,t2) -> if elementRight xs t1 = List.empty then t1::xs else element xs t1

3。当使用相互递归函数时，您需要对所有声明使用and关键字，但第一个是这样的：

let rec elementRight xs = function
  ...
and elementLeft xs = function
  ...
and element xs = function
  ...

将树转换为列表

1 个答案: