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2.64.rkt
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2.64.rkt
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#lang sicp
;; Exercise 2.64:
;; The following procedure list->tree converts an ordered list to a balanced binary tree.
;; The helper procedure partial-tree takes as arguments an integer n and list of at least
;; n elements and constructs a balanced tree containing the first n elements of the list.
;; The result returned by partial-tree is a pair (formed with cons) whose car is the constructed tree and
;; whose cdr is the list of elements not included in the tree.
(define (entry tree) (car tree))
(define (left-branch tree) (cadr tree))
(define (right-branch tree) (caddr tree))
(define (make-tree entry left right)
(list entry left right))
(define (list->tree elements)
(car (partial-tree
elements (length elements))))
(define (partial-tree elts n)
(if (= n 0)
(cons '() elts)
(let ((left-size
(quotient (- n 1) 2)))
(let ((left-result
(partial-tree
elts left-size)))
(let ((left-tree
(car left-result))
(non-left-elts
(cdr left-result))
(right-size
(- n (+ left-size 1))))
(let ((this-entry
(car non-left-elts))
(right-result
(partial-tree
(cdr non-left-elts)
right-size)))
(let ((right-tree
(car right-result))
(remaining-elts
(cdr right-result)))
(cons (make-tree this-entry
left-tree
right-tree)
remaining-elts))))))))
;; Tests
(partial-tree '(1 3 5 7 9 11) 6)
;; Write a short paragraph explaining as clearly as you can how partial-tree works.
;; cut the elts from middle, use the middle element as an entry, and do the same thing for left and right branch
;; Draw the tree produced by list->tree for the list (1 3 5 7 9 11).
;; 5
;; / \
;; 1 9
;; \ / \
;; 3 7 11
;;
;; What is the order of growth in the number of steps required by list->tree to convert a list of n elements?
;; O(n)