让以下成为树拉链:
type 'a ntree = Node of 'a * 'a ntree list
type 'a context =
| Top
| Context of 'a ntree list * 'a * 'a context * 'a ntree list
type 'a tree_zipper = TZ of 'a context * 'a ntree
let tz1 = TZ (Context (
[Node ("*", [Node ("a", []); Node ("b", [])])], (*left siblings in reverse order *)
"+", (*value of parent node *)
Top, (*parent's context *)
[]), (*right sibling in order *)
Node ("*", [Node ("c", []); Node ("d", [])]));;
下一个操作(我的)有什么区别:
let next_exn (TZ (c, t)) = (* on the right, not possible when context is top or right = []*)
match c with
| Top -> failwith "right of top node"
| Context (left, v, up, []) -> failwith "right of last child" (*explain*)
| Context (left, v, up, r::right) -> TZ (Context (t::left, v, up, right), r)
和提供的解决方案:
let next_exn (TZ (c, t)) = (* on the right, not possible when context is top or right = []*)
match c with
| Top -> failwith "right of top node"
| Context(left, v, up, r::right) -> TZ (Context(t::left, v, up, right), r)
| _ -> failwith "right of last child"
我假设它们是一样的,但我可能会遗漏一些东西。
另外,为什么我们将
down_exn
中的左兄弟列表设置为[]
中的:
let down_exn (TZ (c, t)) =
match t with (* go to first child *)
| Node (_, []) -> failwith "down of leaf"
| Node (v, d::ds) -> TZ (Context([], v, c, ds), d)
?
是的,两者实际上是等价的。在
match
表达式中重新排序模式时,请记住它将评估匹配的 first 模式的相应表达式。
这是一个见仁见智的问题,但我当然更喜欢在更复杂的结果之前获得最简单的结果(包括异常或递归基础案例)。基于阅读大量代码,我怀疑我并不孤单。
关于你的第二个问题,如果我们“向下”,我们将“向下”到left-most节点,所以左边不能有兄弟节点。