Question

我试图在F＃中构建一个二叉树，但是当我尝试测试我的代码时，我遇到了上面的问题。

这是我的代码：

type TreeNode<'a> = { Key: int; Val: 'a }
type Tree<'a> =  { LT: Tree<'a> option; TreeNode: TreeNode<'a>; RT: Tree<'a> option; }

//insert a node according to Binary Tree operation
let rec insert (node: TreeNode<'a>) (tree: Tree<'a> option) =
    match tree with
    | None -> {LT = None; RT = None; TreeNode = node }
    | Some t when node.Key < t.TreeNode.Key -> insert node t.LT
    | Some t when node.Key > t.TreeNode.Key -> insert node t.RT

let t = seq { for i in 1 .. 10 -> { Key = i; Val = i } }|> Seq.fold (fun a i -> insert i a) None

Answer 1

您的insert功能需要option<Tree<'T>>，但会返回Tree<'T>。执行fold时，您需要保持相同类型的状态 - 因此，如果要使用None来表示空树，则状态必须是可选类型。

解决此问题的方法是将insert的结果包装在Some中：

let tree = 
  seq { for i in 1 .. 10 -> { Key = i; Val = i } }
  |> Seq.fold (fun a i -> Some(insert i a)) None

Answer 2

我现在已经解决了......它应该如下所示：

＆＃xA;＆＃xA;

 类型TreeNode＆lt;'a＆gt; = {Key：int; Val：'a}＆＃xA;输入树＆lt;'a＆gt; = {TreeNode：TreeNode＆lt;'a＆gt ;; RT：树＆lt;'a＆gt;选项; LT：树＆lt;'a＆gt;选项; }＆＃xA;＆＃xA; //根据二叉树操作插入节点＆＃xA;让rec插入（节点：TreeNode＆lt;'a＆gt;）（树：树＆lt;'＆gt;选项）=＆＃xA;匹配树＆＃xA; |无 - ＆gt; {LT =无; RT =无; TreeNode = node}＆＃xA; |当node.Key＆lt; t.TreeNode.Key  - ＆gt; {TreeNode = t.TreeNode; LT =一些（插入节点t.LT）; RT = t.RT}＆＃xA; |当node.Key＆gt;时有些t.TreeNode.Key  - ＆gt; {TreeNode = t.TreeNode; RT = Some（插入节点t.RT）; LT = t.LT}＆＃xA;＆＃xA;令t = seq {for i in 1 ... 10-＆gt; {Key = i; Val = i}} |＆gt; Seq.fold（有趣的i  - ＆gt; Some（插入i））无＆＃xA;

＆＃xA;

F＃ - 预计是选项但有类型

2 个答案: