#| BEGIN (Введите свое решение) |#
(define (union-set s1 s2)
  (list->tree (union-set-list
                              (tree->list-2 s1)
                              (tree->list-2 s2))))

(define (intersection-set s1 s2)
  (list->tree (intersection-set-list 
                                     (tree->list-2 s1)
                                     (tree->list-2 s2))))

(define (intersection-set-list set1 set2)
  (cond ((or (null? set1) (null? set2)) '())
        ((element-of-set-list? (car set1) set2)
         (cons (car set1)
               (intersection-set-list (cdr set1) set2)))
        (else (intersection-set-list (cdr set1) set2))))

(define (element-of-set-list? x set)
  (cond ((null? set) #f)
        ((equal? x (car set)) #t)
        (else (element-of-set-list? x (cdr set)))))

(define (adjoin-set x set)
  (cond ((element-of-set-list? x set) set)
        ((null? set) (cons x set))
        ((> x (car set)) (cons (car set) (adjoin-set x (cdr set))))
        ((< x (car set)) (cons x set))))

(define (union-set-list set1 set2)
  (cond [(null? set1) set2]
        [(element-of-set-list? (car set1) set2) (union-set-list (cdr set1) set2)]
        [else (union-set-list (cdr set1) (adjoin-set (car set1) set2))]))
#| END |#