将树转换为列表

Question

如何将树转换为列表： 到目前为止，我有以下代码，但它给出了几个问题：

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

你的代码有很多问题：

1. 你不应该在行尾有;;（我猜这意味着你正在复制并粘贴到repl中，你应该真正使用fsx文件并使用“发送到repl ")。

2. 这个：| LF ->false 返回一个布尔值，而这个：| Br(m,t1,t2) -> if elementRight xs t1 = false then t1::xs else element xs t1 返回一个'a list。一个表达式只能有一种类型，所以返回两种是编译错误。我猜你真正的意思是让叶子返回[]，然后在你的分支案例中检查空列表，如下所示：

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

3 .使用相互递归函数时，您需要对所有声明使用 and 关键字，但第一个声明如下：

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