我有一个可以创建树结构的结构:
(struct node (value left middle right))
和另一个定义叶子节点的结构:
(struct emptyNode ())
我如何制作一个通过将值加在一起将树折叠成一个值的函数,从左子树开始,然后是中间,然后是右?
我正在考虑将树转换为列表,首先是左子树的顺序,然后是中间,然后是右子树的顺序,然后对列表进行折叠,但是不确定如何完成它,或者不确定正确的方法:
(define (treeToList tree)
(cond
[(node? tree) (append (node-left tree)) (treeToList (node-left tree))
(append (node-middle tree)) (treeToList (node-middle tree))
(append (node-right tree)) (treeToList (node-right tree))]
[else ] ;do nothing, leaf node
))
因此,执行此操作的一种好方法是使用Racket的模式匹配。除此之外,使用基于议程的搜索也很不错,这样整个过程都是迭代的。因此,执行求和的函数将具有三个参数:
然后,它根据与这些参数匹配的模式,循环遍历节点并从议程中推送和弹出内容来决定要做什么。
给定node
和empty-node
的定义,以及由它们构成的示例树,如下所示:
(struct node (value left middle right))
(struct empty-node ())
(define sample-tree
(node 1
(node 2
(empty-node)
(node 3 (empty-node) (empty-node) (empty-node))
(node -1
(node 1 (empty-node) (empty-node) (empty-node))
(empty-node)
(empty-node)))
(node 10 (empty-node) (empty-node) (empty-node))
(empty-node)))
我们可以编写一个函数像这样对树求和:
(define (sum-node-tree node-tree)
(define/match (snt-loop thing agenda sum)
[((empty-node) '() s)
;; an empty node, nothing on the agenda: we're done
s]
[((empty-node) (cons next more) s)
;; an empty node, but there is an agenda, so start on the next agenda
;; item
(snt-loop next more s)]
[((node (? number? v) l m r) a s)
;; a node with a value: sum the value into the total, push the middle
;; and right children onto the agenda, and start on the left child.
(snt-loop l (list* m r a) (+ s v))]
[(_ _ _)
;; something bad in the tree
(error 'sum-node-tree "bogus tree")])
(snt-loop node-tree '() 0))
然后
> (sum-node-tree sample-tree)
16