表达式树评估

| 我正在尝试评估表达式树。这是我的代码： 
(define (eval-tree expr-tree)
  (eval-treeaux eval-tree expr-tree))

(define (eval-treeaux f expr-tree)
  (if (null? expr-tree)
      0
      (cond ((equal? \'+ (operator expr-tree)) (+ (left-op expr-tree) (right-op expr-tree)))
            ((equal? \'- (operator expr-tree)) (- (left-op expr-tree) (right-op expr-tree)))
            ((equal? \'* (operator expr-tree)) (* (left-op expr-tree) (right-op expr-tree)))
            ((equal? \'/ (operator expr-tree)) (/ (left-op expr-tree) (right-op expr-tree)))
            (else (f eval-treeaux expr-tree)))))
当我运行这个“ 1”时，它给了我这个错误： 
+: expects type <number> as 1st argument, given: (6 * 3); other arguments 
   were: (4 - 2)
谁能告诉我问题出在哪里以及如何解决？ 这是我的新代码： 
(define (eval-tree expr-tree)
  (if (null? expr-tree)
      0
      ((eval-treeaux eval-tree (car expr-tree)) (eval-tree (cdr expr-tree)))))

(define (eval-treeaux f expr-tree)
  (cond ((null? expr-tree) \'())
        ((not (isExpression expr-tree)) (list expr-tree))
        (else (cond ((equal? \'+ (operator expr-tree)) (+ (left-op expr-tree) (right-op expr-tree)))
                    ((equal? \'- (operator expr-tree)) (- (left-op expr-tree) (right-op expr-tree)))
                    ((equal? \'* (operator expr-tree)) (* (left-op expr-tree) (right-op expr-tree)))
                    ((equal? \'/ (operator expr-tree)) (/ (left-op expr-tree) (right-op expr-tree)))
                    (else (f eval-treeaux expr-tree))))))
但是，当我运行此
(eval-tree \'((6 * 3) + (4 - 2)))
时，出现以下错误： 
procedure application: expected procedure, given: 2; arguments were: 0
我真的很困惑。谁能帮我吗？ 谢谢     
2020-03-10 2 条评论
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    已邀请:

    宠封钞轰

    表达式“ 6”似乎由2个子表达式组成： 
     (6 * 3)  ; e_1
 (4 - 2)  ; e_2
    您需要递归计算子表达式，然后才能评估父表达式。     
    2020-03-10 0 0
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