Exercise 2.31
Abstract your answer to Exercise 2.30 to produce a procedure t r e e − m a p tree-map tree−map with the property that s q u a r e − t r e e square-tree square−tree could be defined as
(define (square-tree tree) (tree-map square tree))
这道题跟上面一道的 map 实现几乎一模一样，我还以为我理解错题目了，上网搜了一下，发现别人也是这么写的，那就这样吧。
(define (tree-map f tree)(map (lambda (sub-tree)(if (pair? sub-tree)(tree-map f sub-tree)(f sub-tree)))tree))(define (square-tree tree) (tree-map square tree))(square-tree(list 1(list 2 (list 3 4) 5)(list 6 7))); 执行结果
'(1 (4 (9 16) 25) (36 49))
