## SICP – cap. 2 – sets come alberi binari – 85 – esercizio Continuo da qui, copio qui.

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?

Qua non ci provo; chiamo Bill the Lizard 💥

The `partial-tree` procedure works by dividing the list into three parts, a center element (the root node of the tree), everything before the center element, and everything after the center element. All the elements before the center element are then passed to a recursive call to `partial-tree` to create the left branch of the tree, and all the elements after the center element are passed recursively to `partial-tree to` create the right branch. These recursive call continue until no elements are remaining, and the balanced binary tree is assembled.

The tree produced by `list->tree` for the list `(1 3 5 7 9 11)` is: To verify this, we can simply call the procedure. Next we’re asked what is the order of growth in the number of steps required by `list->tree` to convert a list of n elements? The procedure only needs to visit each element of the list once, and it only performs a `cons` for each element it visits, so the number of steps is proportional to the size of the list, or `O(n)`.

sicp-ex risponde in modo simile, più conciso.
Risultato simile per Drewiki.
Insomma sono solo io che devo applicarmi di più 😐

🤢

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