Exercise 2.27. Modify your reverse procedure of exercise 2.18 to produce a deep-reverse procedure that takes a list as argument and returns as its value the list with its elements reversed and with all sublists deep-reversed as well. For example,
(define x (list (list 1 2) (list 3 4)))
x
((1 2) (3 4))
(reverse x)
((3 4) (1 2))
(deep-reverse x)
((4 3) (2 1))
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(define nil '()) | |
(define (append list1 list2) | |
(if (null? list1) | |
list2 | |
(cons (car list1) (append (cdr list1) list2)))) | |
(define (reverse list1) | |
(if | |
(null? list1) | |
nil | |
(if | |
(null? (cdr list1)) | |
list1 | |
(append | |
(reverse (cdr list1)) | |
(cons (car list1) nil))))) | |
(define (deep-reverse tree1) | |
(map | |
(lambda (x) | |
(if | |
(list? x) | |
(deep-reverse x) | |
x)) | |
(reverse tree1))) | |
(define x (list (list 1 2) (list 3 4))) | |
(define y (list (list 1 2) (list 3 (list 4 5)))) | |
x | |
(reverse x) | |
(deep-reverse x) | |
y | |
(reverse y) | |
(deep-reverse y) | |
333 | |
(reverse (list 1 2)) | |
(reverse (list 1 2 3 4 5)) | |
(reverse (list 1)) | |
(reverse '()) | |
(deep-reverse (list 1 2)) | |
(deep-reverse (list 1 2 3 4 5)) | |
(deep-reverse (list 1)) | |
(deep-reverse '()) | |