SICP exercise 2.30
From Drewiki
Problem
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)))
Output:
(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.
Solution
Without higher-order procedures:
(define nil (quote ())) (define (square x) (* x x)) (define (square-tree tree) (cond ((null? tree) nil) ((pair? tree) (cons (square-tree (car tree)) (square-tree (cdr tree)))) (else (square tree))))
Test:
(square-tree (list 1 (list 2 (list 3 4) 5) (list 6 7)))
Output:
(1 (4 (9 16) 25) (36 49))
With map:
(define (square-tree tree) (map (lambda (sub-tree) (if (pair? sub-tree) (square-tree sub-tree) (square sub-tree))) tree))
Test:
(square-tree (list 1 (list 2 (list 3 4) 5) (list 6 7)))
Output:
(1 (4 (9 16) 25) (36 49))
