sicp每日一题[2.65]

合集 - sicp(77)
1.sicp每日一题[1.42]2.sicp每日一题[1.43]3.sicp每日一题[1.44]4.sicp每日一题[1.45]5.sicp每日一题[1.46]6.sicp每日一题[2.1]7.sicp每日一题[2.2]8.sicp每日一题[2.3]9.sicp每日一题[2.4]10.sicp每日一题[2.5]11.sicp每日一题[2.6]12.sicp每日一题[2.7]13.sicp每日一题[2.8]14.sicp每日一题[2.9]15.sicp每日一题[2.11]16.sicp每日一题[2.10]17.sicp每日一题[2.12]18.sicp每日一题[2.13-2.16]19.sicp每日一题[2.17]20.sicp每日一题[2.18]21.sicp每日一题[2.19]22.sicp每日一题[2.20]23.sicp每日一题[2.21]24.sicp每日一题[2.22-2.23]25.sicp每日一题[2.24-2.27]26.sicp每日一题[2.28]27.sicp每日一题[2.29]28.sicp每日一题[2.30]29.sicp每日一题[2.31]30.sicp每日一题[2.32]31.sicp每日一题[2.33]32.sicp每日一题[2.34]33.sicp每日一题[2.35]34.sicp每日一题[2.36-2.37]35.sicp每日一题[2.38-2.39]36.sicp每日一题[2.40]37.sicp每日一题[2.41]38.sicp每日一题[2.42]39.sicp每日一题[2.43]40.sicp每日一题[2.44]41.sicp每日一题[2.45]42.sicp每日一题[2.46]43.sicp每日一题[2.47]44.sicp每日一题[2.48]45.sicp每日一题[2.49]46.sicp每日一题[2.50]47.sicp每日一题[2.51]48.sicp每日一题[2.52]49.sicp每日一题[2.53-2.55]50.sicp每日一题[2.56]51.sicp每日一题[2.57]52.sicp每日一题[2.58]53.sicp每日一题[2.59]54.sicp每日一题[2.60]55.sicp每日一题[2.61]56.sicp每日一题[2.62]57.sicp每日一题[2.63-2.64]
58.sicp每日一题[2.65]
59.sicp每日一题[2.66]60.sicp每日一题[2.67-2.68]61.sicp每日一题[2.69]62.sicp每日一题[2.70]63.sicp每日一题[2.71]64.sicp每日一题[2.72]65.sicp每日一题[2.73]66.sicp每日一题[2.74]67.sicp每日一题[2.75]68.sicp每日一题[2.76]69.sicp每日一题[2.77]70.sicp每日一题[2.78]71.sicp每日一题[2.79]72.sicp每日一题[2.80]73.sicp每日一题[2.81]74.sicp每日一题[2.82]75.sicp每日一题[2.83]76.sicp每日一题[2.84]77.sicp每日一题[2.85-2.86]
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Exercise 2.65

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

这道题难度还是挺大的，一开始我不知道怎么直接对2颗树求交集和并集，而且要在 O(n) 的时间复杂度内。上网看了一下别人的答案，发现他们都是先用 2.63 的 tree->list-2 把树转为列表，然后再求交集并集，最后再用 2.64 的 list->tree 把结果转成树结构。
这样我就明白怎么做了，尤其是求并集可以直接参考 2.62 的答案，但是稍作修改，让它能去除重复元素，求交集虽然没有现成的，但是参考并集的方法也不难。

(define (union-tree tree1 tree2) 
  (define (union-set set1 set2) 
    (cond ((null? set2) set1) 
          ((null? set1) set2)
          (else (let ((s1 (car set1))
                      (s2 (car set2)))
                  (cond ((= s1 s2) (cons s1 (union-set (cdr set1) (cdr set2))))
                        ((< s1 s2) (cons s1 (union-set (cdr set1) set2)))
                        (else (cons s2 (union-set set1 (cdr set2)))))))))
  (list->tree (union-set (tree->list-2 tree1) (tree->list-2 tree2))))
  
(define (intersection-tree tree1 tree2) 
  (define (intersection-set set1 set2) 
    (if (or (null? set1) (null? set2))
        '()
        (let ((s1 (car set1))
              (s2 (car set2)))
          (cond ((= s1 s2) (cons s1 (intersection-set (cdr set1) (cdr set2))))
                ((< s1 s2) (intersection-set (cdr set1) set2))
                (else (intersection-set set1 (cdr set2)))))))
  (list->tree (intersection-set (tree->list-2 tree1) (tree->list-2 tree2)))) 


(define tree1 (list->tree (list 1 3 5 7 9 11)))
(define tree2 (list->tree (list 2 4 6 7 8 9 10)))

(intersection-tree tree1 tree2)
(union-tree tree1 tree2)

; 结果如下
'(7 () (9 () ()))
'(6 (3 (1 () (2 () ())) (4 () (5 () ()))) (9 (7 () (8 () ())) (10 () (11 () ()))))
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