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.

; (i) Without using higher-order procedure
(define (square-tree tree)
    (cond ((null? tree) tree)
          ((not (pair? tree)) (square tree))
          (else (cons (square-tree (car tree))
                      (square-tree (cdr tree))))))
(square-tree
    (list 1
        (list 2 (list 3 4) 5)
        (list 6 7)))
;Value 2: (1 (4 (9 16) 25) (36 49))

; (ii) Using `map`
(define (square-tree tree)
    (map (lambda (sub-tree)
            (if (pair? sub-tree)
                (square-tree sub-tree)
                (square sub-tree)))
         tree))
(square-tree
    (list 1
        (list 2 (list 3 4) 5)
        (list 6 7)))
;Value 3: (1 (4 (9 16) 25) (36 49))