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# SICP exercise 2.65

## Problem

Use the results of exercises 2.63 and 2.64 to give ￼θ(n) implementations of union-set and intersection-set for sets implemented as (balanced) binary trees.

## Solution

```(define (entry tree) (car tree))

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

(define (tree->list 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 '()))

(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))))))))

(define (union-set set1 set2)
(define (union-set-list list1 list2)
(cond ((null? list1) list2)
((null? list2) list1)
(else (let ((x1 (car list1))
(x2 (car list2)))
(cond ((= x1 x2)
(cons x1 (union-set-list (cdr list1) (cdr list2))))
((< x1 x2)
(cons x1 (union-set-list (cdr list1) list2)))
(else
(cons x2 (union-set-list list1 (cdr list2)))))))))
(list->tree (union-set-list (tree->list set1)
(tree->list set2))))

(define (intersection-set set1 set2)
(define (intersection-set-list list1 list2)
(if (or (null? list1) (null? list2))
'()
(let ((x1 (car list1)) (x2 (car list2)))
(cond ((= x1 x2)
(cons x1
(intersection-set-list (cdr list1)
(cdr list2))))
((< x1 x2)
(intersection-set-list (cdr list1) list2))
((< x2 x1)
(intersection-set-list list1 (cdr list2)))))))
(list->tree (intersection-set-list (tree->list set1)
(tree->list set2))))```

Tests:

```(define s1 (list->tree '(1 2 3 5 6 8 10 11)))
(define s2 (list->tree '(2 4 6 8 11 13)))
(tree->list (union-set s1 s2))```

Output:

```(1 2 3 4 5 6 8 10 11 13)
```
`(tree->list (intersection-set s1 s2))`

Output:

```(2 6 8 11)
```

My solutions transform the tree to a sorted list, perform the algorithms developed in the text and in exercise 2.62, and then transform the sorted list back into a tree.

Each procedure used by union-set or intersection-set to perform the operations is θ(n). Except for the tree->list transformation, which is called twice per application of union-set or intersection-set, each procedure is called once per application, so those implementations are also θ(n).