計算機程序的構造和解釋 練習題2.63

(a)兩個方法運行結果一致，兩個方法都是遞歸的方法把樹結構編程列表，不同的地方在於，第一個方法用的append連接左右樹，而第二個用的是cons連接。
(b)tree->list-1會執行N次append和cons，而tree->list-2，只會執行n次cons，append的時間複雜度是θ(n),cons的時間複雜度是θ(1)，所以tree->list-1和tree->list-2的時間複雜度分別爲爲θ(n^2)和 θ(n)，tree->list-2過程的步數會增長慢點。

#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 (tree->list-1 tree)
  (if (null? tree)
      '()
      (append (tree->list-1 (left-branch tree))
              (cons (entry tree)
                    (tree->list-1 (right-branch tree))))))

(define (tree->list-2 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 a (make-tree 7
                     (make-tree 3
                                (make-tree 1 null null)
                                (make-tree 5 null null))
                     (make-tree 9
                                null 
                                (make-tree 11 null null))))
(define b (make-tree 3
                     (make-tree 1 null null)
                     (make-tree 7
                                (make-tree 5 null null)
                                (make-tree 9
                                           null
                                           (make-tree 11 null null)))))

(define c (make-tree 5
                     (make-tree 3
                                (make-tree 1 null null)
                                null)
                     (make-tree 9
                                (make-tree 7  null null)
                                (make-tree 11 null null))))
(tree->list-1 a)
(tree->list-2 a)

(tree->list-1 b)
(tree->list-2 b)

(tree->list-1 c)
(tree->list-2 c)

運行結果

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