在方案中按顺序遍历树

Question

我想按顺序遍历节点包含整数的树，如果所有数字都按顺序排列（即下一个数字大于或等于前一个数字），则返回 #t。

到目前为止我的尝试

 (define (in-order tree number)  
    (cond [(leaf? tree) (>= leafnumber  number)]
          [(cons? leftsubtree) (in-order leftsubtree leftnumber)]
          [(cons? rightsubtree) (in-order rightsubtree rightnumber)]))

当我切换分支时，不知何故我的数字参数似乎是错误的。 我需要使用另一个变量来存储这些吗？

Answer 1

这是我的建议：

; in-order : tree number -> number or #f
;   if all numbers in the tree are larger than number, then return the maximum number in tree,
;   otherwise 
(define (in-order tree number)
  (cond
    [(leaf? tree)   (if (>= leafnumber  number)
                        leafnumber               ; leafnumber is the largest number in tree
                        #f)]                     ; not all numbers in tree are smaller than number
    [else           (let ([leftnumber (in-order  leftsubtree number)])
                      (let ([rightnumber (in-order rightsubtree leftnumber)])
                        (and leftnumber
                             rightnumber)))]))

(in-order tree -inf.0)