Exercise 2.30. Define a procedure square-tree analogous to the square-list procedure of exercise 2.21. That is, square-list 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.
仿照课本中的 scale-tree 即可。
(define (square-tree tree)
(map (lambda (subtree)
(if (pair? subtree)
(square-tree subtree)
(* subtree subtree)))
tree))
(define (square-tree-iter tree)
(cond ((null? tree) '())
((not (pair? tree)) (* tree tree))
(else
(cons (square-tree-iter (car tree))
(square-tree-iter (cdr tree))))))
;;; tests begin
(load "../testframe.scm")
(define tree '(1 2 (3 4 5) (6 (7 (8 (9))))))
(assertequal? (square-tree tree) '(1 4 (9 16 25) (36 (49 (64 (81))))))
(assertequal? (square-tree-iter tree) (square-tree tree))