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SICP exercise 2.31

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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)))


 (list 1
       (list 2 (list 3 4) 5)
       (list 6 7)))


(1 (4 (9 16) 25) (36 49))
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