Ex-2.30

;without using any higher order procedures
(define (square-tree tree)
  (cond
   ((null? tree) '())
   ((list? (car tree))
    (append (list (square-tree (car tree)))
            (square-tree (cdr tree))))
   (else
    (cons (square (car tree))
          (square-tree (cdr tree))))))

;using higher order procedures
(define (square-tree tree)
  (map
   (lambda (x)
     (if (pair? x)
         (square-tree x) (square x))) tree))