計算機程序的構造和解釋 練習題2.64

#lang racket
(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-set? x set)
  (cond ((null? set) false)
        ((= x (entry set)) true)
        ((< x (entry set))
         (element-of-set? x (left-branch set)))
        ((> x (entry set))
         (element-of-set? x (right-branch set)))))
(define (adjoin-set x set)
  (cond ((null? set) (make-tree x '() '()))
        ((= x (entry set)) set)
        ((< x (entry set))
         (make-tree (entry set)
                    (adjoin-set x (left-branch set))
                    (right-branch set)))
        ((> x (entry set))
         (make-tree (entry set)
                    (left-branch set)
                    (adjoin-set x (right-branch set))))))

(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))))))))
(list->tree '(1 3 5 7 9 11))

運行結果

'(5 (1 () (3 () ())) (9 (7 () ()) (11 () ())))

(a)遞歸的思路是將一個列表的節點分爲兩部分，然後前面的(n-1)/2個元素組成左邊的樹，中間的元素是中間節點，後面的n-left個數加1組成右邊樹的元素，然後遞歸，直到n=0時，返回’()空集合，不好理解的話，可以按每一部調試程序去執行，就能清楚地看到整個的調用過程了。remaining-elts其實就是每層樹遺留下來的元素，也就是右邊樹和中間數值組成的列表。
調用步驟羅列：

(partial-tree (1 3 5 7 9 11) 7)
(make-tree 5 (partial-tree (1 3)) (partial-tree (7 9 11) 3))
//左半部分
(partial-tree (1 3) 3)
(make-tree 1 (null (partial-tree (3) 1))
(partial-tree (3) 1)
(make-tree 3 (() ()))
//右半部分
(partial-tree (7 9 11) 3)
(make-tree 9 ((partial-tree (7) 1)(partial-tree (11) 1))
(partial-tree (7) 1)
(make-tree 7 (() ()))
(partial-tree (11) 1)
(make-tree 11 (() ()))

下面是最終的樹圖

(b)
接着按上面（a）小題的分解步驟，一共是6個元素，最終執行了6次make-tree，依次類推n個元素應該也是執行n次make-tree，時間複雜度應該爲θ(n)。