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sicp-all-tasks/sicp/2_002e64

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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 (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))))))))
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).
What is the order of growth in the number of steps required by
list->tree to convert a list of
n
elements?