sicp每日一題[2.30]

Exercise 2.30

Define a procedure square-tree analogous to the square-list procedure of Exercise 2.21. That is, square-tree should behave as follows:

(square-tree
 (list 1
       (list 2 (list 3 4) 5)
       (list 6 7)))
(1 (4 (9 16) 25) (36 49))

Define square-tree both directly (i.e., without using any higher-order procedures) and also by using map and recursion.

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這個題目沒啥難度，直接仿照 scale-tree 函式，修改一下函式名就行了。

(define (square-tree tree)
  (cond ((null? tree) nil)
        ((not (pair? tree)) (square tree))
        (else (cons (square-tree (car tree))
                    (square-tree (cdr tree))))))

(define (square-tree-by-map tree)
  (map (lambda (sub-tree)
         (if (pair? sub-tree)
             (square-tree-by-map sub-tree)
             (square sub-tree)))
       tree))


(define test (list 1
                   (list 2 (list 3 4) 5)
                   (list 6 7)))

(square-tree test)
(square-tree-by-map test)

; 執行結果
'(1 (4 (9 16) 25) (36 49))
'(1 (4 (9 16) 25) (36 49))