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SICP 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))
Here's one way to do it:
(define nil (quote ())) (define (square x) (* x x)) (define (tree-map f tree) (map (lambda (sub-tree) (if (pair? sub-tree) (tree-map f sub-tree) (f sub-tree))) tree))
(square-tree (list 1 (list 2 (list 3 4) 5) (list 6 7)))
(1 (4 (9 16) 25) (36 49))