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 (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))))))))
; (a) Write a short paragraph explaining as clearly as you can how partial-tree
; works. Draw the tree produced by list->tree for the list (1 3 5 7 9 11).
; Procedure `partial-tree` will construct a tree of `elts`, where root is the middle
; element (median) and left-subtree is `partial-tree` of elements less than middle
; element (this-entry) and right-substree is `partial-tree` of elements greater than
; `this-entry`
; For (1 3 5 7 9 11), it constructs the following tree
; 5
; / \
; 1 9
; \ / \
; 3 7 11
; (b) What is the order of growth in the number of steps required by list->tree to
; convert a list of n elements?
; T(n) = 2T(n/2) + O(1)
; T(n) = O(n)