計算機程序的構造和解釋 練習題2.65

可以用之前的有序表的過程union-set，intersection-set以及平衡樹的轉化過程tree->list、list->tree來寫。
這幾個過程的複雜度都是θ(n)，最終過程裏面沒有迭代或者遞歸，所以複雜度還是θ(n)。

(define (union-tree set1 set2)
   (list->tree (union-set (tree->list set1) (tree->list set2))))
(define (intersection-tree set1 set2)
   (list->tree (intersection-set (tree->list set1) (tree->list set2))))

附上完整的過程和運行結果

#lang racket
(define (element-of-set? x set)
  (cond ((null? set) #f)
        ((= x (car set)) #t)
        ((< x (car set)) #f)
        (else (element-of-set? x (cdr set)))))
(define (adjoin-set x set) 
    (cond ((null? set) (list x))
          ((= x (car set)) set)
          ((< x (car set)) (cons x set))
          (else (cons (car set) (adjoin-set x (cdr set))))))

(define (intersection-set set1 set2)
  (if (or (null? set1) (null? set2)) 
      '()
      (let ((x1 (car set1)) (x2 (car set2)))
        (cond ((= x1 x2)
               (cons x1 (intersection-set (cdr set1)
                                          (cdr set2))))
              ((< x1 x2)
               (intersection-set (cdr set1) set2))
              ((< x2 x1)
               (intersection-set set1 (cdr set2)))))))

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

(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))

(define (element-of-tree? x set)
  (cond ((null? set) false)
        ((= x (entry set)) true)
        ((< x (entry set))
         (element-of-tree? x (left-branch set)))
        ((> x (entry set))
         (element-of-tree? x (right-branch set)))))
(define (adjoin-tree x set)
  (cond ((null? set) (make-tree x '() '()))
        ((= x (entry set)) set)
        ((< x (entry set))
         (make-tree (entry set)
                    (adjoin-tree x (left-branch set))
                    (right-branch set)))
        ((> x (entry set))
         (make-tree (entry set)
                    (left-branch set)
                    (adjoin-tree x (right-branch set))))))
(define (union-tree set1 set2)
   (list->tree (union-set (tree->list set1) (tree->list set2))))
(define (intersection-tree set1 set2)
   (list->tree (intersection-set (tree->list set1) (tree->list set2))))

(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 a (list->tree '(1 3 5 7 9 11)))
(define b (list->tree '(4 6 7 8 9 10)))
(union-tree a b)
(tree->list (union-tree a b))

運行結果

'(6 (3 (1 () ()) (4 () (5 () ()))) (9 (7 () (8 () ())) (10 () (11 () ()))))
'(1 3 4 5 6 7 8 9 10 11)