我试图通过使用折叠函数来获得树的产品，到目前为止，这就是我所拥有的。我很困惑如何在遍历树的时候使用fold方法 数据类型'a bin_tree ='a | 'a bin_tree *'的节点a bin_tree 好玩的treefold g z空白= z | treefold gz（Node（l，x，r））= g（x，g（treefold gzl，treefold gzr）
解决方案
在折叠二叉树时，
datatype'a tree = Leaf |'a tree *'a *'a tree的分支您可以在避免堆栈溢出s。
对于涉及遍历树的一些问题，提供遍历的上下文可能很有用，如
fun treecata_preorder f acc1 Leaf = acc1 | treecata_preorder f acc1（branch as branch（left，a，right））= let val acc2 = treecata_preorder f acc1 left val acc3 = f（a，branch，acc2） val acc4 = treecata_preorder f acc3 right acc $ 4中的结尾这是 treefold_preorder ，其中 f 被送到整个分支。
这可以让你例如找到一个祖先树中的一个谓词持有它们的子树，
fun treefilter pred = treecata_preorder（fn （x，xtree，acc）=> if pred xtree then x :: acc else acc）[] fun branchValue Leaf = NONE | branchValue（Branch（_，value，_））=某个值 有趣的父母叶= [] |父类（Branch（left，_，right））= List.mapPartial（fn xopt => xopt）[branchValue left，branchValue right] type name = string 类型age = int datatype person =名字的年龄*年龄 有趣的退休（Person（_，age））= age> = 70 fun hasRetiredParent tree = List。存在退休（父母树） val personsWith RetiredParents = treefilter hasRetiredParent树遍历的另一个概念拉链（LYAH章节）。
I am trying to get the product of a tree by using the fold function so far this is what I have. I am confused on how to use the fold method while transversing the tree
datatype 'a bin_tree = Leaf of 'a | Node of 'a bin_tree * 'a bin_tree fun treefold g z Empty = z | treefold g z (Node (l, x, r)) = g(x, g(treefold g z l, treefold g z r)解决方案
When folding a binary tree,
datatype 'a tree = Leaf | Branch of 'a tree * 'a * 'a treeyou may traverse it in different ways. Among common strategies you have,
(* Pre-order *) fun treefold_preorder f acc1 Leaf = acc1 | treefold_preorder f acc1 (Branch (left, a, right)) = let val acc2 = treefold_preorder f acc1 left val acc3 = f (a, acc2) val acc4 = treefold_preorder f acc3 right in acc4 end (* In-order *) and treefold_inorder f acc1 Leaf = acc1 | treefold_inorder f acc1 (Branch (left, a, right)) = let val acc2 = f (a, acc1) val acc3 = treefold_inorder f acc2 left val acc4 = treefold_inorder f acc3 right in acc4 end (* Post-order *) and treefold_postorder f acc1 Leaf = acc1 | treefold_postorder f acc1 (Branch (left, a, right)) = let val acc2 = treefold_postorder f acc1 left val acc3 = treefold_postorder f acc2 right val acc4 = f (a, acc3) in acc4 endwhich Wikipedia nicely illustrates as,
Usage val treelist = treefold op:: [] val treeproduct = treefold op* 1 val treecount = treefold (fn (_, count) => count + 1) 0Extra
In-order traversal isn't meaningful if each branch/node doesn't have an 'a value.
See also how to apply tail-recursion on trees to avoid stack overflows.
For some problems that involve tree traversal, it may be useful to supply the context of the traversal like paramorphisms do:
fun treecata_preorder f acc1 Leaf = acc1 | treecata_preorder f acc1 (branch as Branch (left, a, right)) = let val acc2 = treecata_preorder f acc1 left val acc3 = f (a, branch, acc2) val acc4 = treecata_preorder f acc3 right in acc4 endThis is a slight generalisation of treefold_preorder in which f is fed the entire branch.
This lets you e.g. find people in an ancestry tree for which a predicate holds for their subtree,
fun treefilter pred = treecata_preorder (fn (x, xtree, acc) => if pred xtree then x::acc else acc) [] fun branchValue Leaf = NONE | branchValue (Branch (_, value, _)) = SOME value fun parents Leaf = [] | parents (Branch (left, _, right)) = List.mapPartial (fn xopt => xopt) [branchValue left, branchValue right] type name = string type age = int datatype person = Person of name * age fun retired (Person (_, age)) = age >= 70 fun hasRetiredParent tree = List.exists retired (parents tree) val personsWithRetiredParents = treefilter hasRetiredParentAnother neat notion for tree traversal are zippers (LYAH chapter).
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