Question

为您提供以下二进制尝试的定义

(define-struct leaf ())
;; interpretation: represents a leaf on a BT, a node with no children

(define-struct node (word left right))
;; interpretation: represents a node on a BT with a word, a left and a right 
;; subtree

;; A BinaryTree (BT) is one of 
;; - (make-leaf)
;; - (make-node String BT BT)

设计程序bt-＆gt; los，它使用BT树并将树的元素作为字符串列表返回。在每个节点上你的功能应该

1.处理左子树

2.处理此节点的单词

3.处理正确的子树

;; bt->los: tree -> los 
;; consumes a BT tree and returns the elements of the tree as a list of strings.
(define (bt->los tree)
(cond 
   [(leaf? tree) ""]
   [(node? tree) 
    (append  (bt->los(list (node-left tree)))
             (list (node-word tree))
             (bt->los (list (node-right tree))))]))

我被困在这里。应该错过了什么。我们不需要递归吗？

输出应该像，

 (define bt-abc (make-node "b" 
                       (make-node "a" (make-leaf) (make-leaf))
                       (make-node "c" (make-leaf) (make-leaf))))
  (bt->los bt-abc) (cons "a" (cons "b" (cons "c" empty)))

Answer 1

你离这儿很近。只是几个错误。

(define (bt->los tree)
 (cond 
   [(leaf? tree) empty]
   [(node? tree) 
    (append  (bt->los (node-left tree))
             (list (node-word tree))
             (bt->los (node-right tree)))]))

首先，您正在构建一个字符串列表。因为它是一个列表，所以你的基本情况应该是empty。不是""。其次，每个节点已经代表一个BT，不需要list它。 bt->los返回一个列表。通过这些简单的更改，它可以按预期的方式运行测试用例。

scheme使用BT树并将树的元素作为字符串列表返回

1 个答案: