Code

#lang racket

; set as binary tree 2.63

(define (entry tree) (car tree))

(define (make-tree entry left right)
(list entry left right))

; converters

(define (tree->list-2 tree)
(define (copy-to-list tree result-list)
(if (null? tree)
result-list
(copy-to-list (left-branch tree)
(cons (entry tree)
(copy-to-list (right-branch tree)
result-list)))))
(copy-to-list tree '()))

; transform ordered list to balanced binary 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 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

(list->tree '(1 3 5 7 9 11))


Output

'(7 (3 (1 () ()) (5 () ())) (11 (9 () ()) ()))


a) partial-tree divides the ordered list in two approximately equal halfs and one entry between them and recursively builds a tree of these parts.

b) $$T(n)=2T(n/2)+O(1)=O(n)$$