sicp每日一題[2.66]

再思即可發表於2024-11-02
每日一題

Exercise 2.66

Implement the lookup procedure for the case where the set of records is structured as a binary tree, ordered by the numerical values of the keys.

這道題還是挺簡單的，由於是有序的二叉樹，所以可以先比較根與要查詢的數字的大小，如果相等就查到了；如果根小於要查詢的數，則遞迴呼叫判斷右子樹中是否包含這個數; 如果根大於要查詢的數，則去左子樹去查即可。

(define (lookup given-key set-of-records)
  (if (null? set-of-records)
      false
      (let ((root (entry set-of-records)))
        (cond ((= given-key root) true)
              ((< given-key root) (lookup given-key (left-branch set-of-records)))
              (else (lookup given-key (right-branch set-of-records)))))))


(define test (list->tree (list 1 2 3 4 5 6 7 8 9)))
(lookup 7 test)
(lookup 10 test)

; 結果如下
#t
#f

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