sicp每日一题[2.30]
Exercise 2.30
Define a procedure s q u a r e − t r e e square-tree square−tree analogous to the s q u a r e − l i s t square-list square−list procedure of Exercise 2.21. That is, s q u a r e − t r e e square-tree 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 s q u a r e − t r e e square-tree square−tree both directly (i.e., without using any higher-order procedures) and also by using map and recursion.
这个题目没啥难度，直接仿照 scale-tree 函数，修改一下函数名就行了。
(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); 执行结果
'(1 (4 (9 16) 25) (36 49))
'(1 (4 (9 16) 25) (36 49))