data Seq a = Nil | Cons a (Seq (a,a))
好的,这是一段声明二叉树嵌套数据类型的Haskell代码,
在OCaml中是否有相应的代码呢?
如果有,请用OCaml代码展示。
我已经尝试过,但我想确认这是否与上面的想法相同:
type tree = Leaf of int | Node of int * tree * tree;;
let l = Leaf 3;;
let a = Node (1, l, l);;
let a = Node (1, a, l);;
let a = Node (1, a, l);;
let a = Node (1, a, l);;
let a = Node (1, a, l);;
let rec value tree = match tree with
| Leaf x -> x
| Node (v, x, y) -> v + (value x) + (value y);;
let rec len tree = match tree with
| Leaf x -> 1
| Node (v, x, y) -> 1 + (len x) + (len y);;
value a;;
len a;;
# #use
"1.ml";;
type tree = Leaf of int | Node of int * tree * tree
val l : tree = Leaf 3
val a : tree = Node (1, Leaf 3, Leaf 3)
val a : tree = Node (1, Node (1, Leaf 3, Leaf 3), Leaf 3)
val a : tree = Node (1, Node (1, Node (1, Leaf 3, Leaf 3), Leaf 3), Leaf 3)
val a : tree =
Node (1, Node (1, Node (1, Node (1, Leaf 3, Leaf 3), Leaf 3), Leaf 3),
Leaf 3)
val a : tree =
Node (1,
Node (1, Node (1, Node (1, Node (1, Leaf 3, Leaf 3), Leaf 3), Leaf 3),
Leaf 3),
Leaf 3)
val value : tree -> int = <fun>
val len : tree -> int = <fun>
- : int = 23
- : int = 11
Cons(“ s”,Nil)
无法转换为tree
。反过来,更有趣的是,树Node(1,Leaf 2,Node(3,Leaf 4,Leaf 5))
没有表示为seq
,因为它是不平衡的-强制平衡是seq
类型的目的。 - Andreas Rossberg