Exercise 2.31. Abstract your answer to exercise 2.30 to produce a procedure tree-map with the property that square-tree could be defined as
(define (square-tree tree) (tree-map square tree))
在 练习 2.31 中我们已经定义了一个使用高阶函数的 square-tree。这里只要替换其中使用到的函数就可以。
(define (tree-map proc tree)
(map (lambda (subtree)
(if (pair? subtree)
(tree-map proc subtree)
(proc subtree)))
tree))
(define (square-tree tree)
(tree-map square 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))))))