# SICP Exercise 2.64

Code

```
#lang racket
; set as binary tree 2.63
(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))
; 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) \(T(n)=2T(n/2)+O(1)=O(n)\)