Question

定义接收树作为参数并返回的反向树方法投资了所有分支机构：

我这样做：

(define (reverse-tree t)
  (map (lambda (x)
         (if (list? x)
             (deep-reverse x)
             x))
       (reverse t)))


 (define tree '(40 (28 (9) (32)) (70 (52) (102))))


(reverse-tree tree)

我退回了这个：

(((102) (52) 70) ((32) (9) 28) 40)

但是我需要你返回反转的节点：

(40 (70 (102) (52)) (28 (32) (9)))

如何回到逆转的点头

Answer 1

对于BST想象一下，您有树的程序。 (make-tree value left right)，tree-value，tree-left和treee-right

(define (reverse-tree tree)
  (if (tree-null? tree)
      tree-null
      (make-tree (tree-value tree)
                 (reverse-tree (tree-right tree))
                 (reverse-tree (tree-left tree)))))

对于一个可以有两个以上孩子的树，我猜你会在每个孩子身上使用它并反转结果。

(define (reverse-tree tree)
  (if (tree-null? tree)
      tree-null
      (apply make-tree 
             (tree-value tree)
             (map reverse-tree (reverse (tree-children tree))))))

;; heres the documentation of the tree
(define (make-tree value . children)
  (cons value children))

(define tree-children cdr)
(define tree-value car)
(define tree-null? null?)
(define tree-null '())

Answer 2

假设一个节点有0或2个分支，这里有一个简单的解决方案：

(define (reverse-tree tree)
  (if (null? tree)
      null
      (let ((root (car tree)) (tree (cdr tree)))
        (if (null? tree)
            (list root)
            (list root (reverse-tree (cadr tree)) (reverse-tree (car tree)))))))

测试

> (reverse-tree '(40 (28 (9) (32)) (70 (52) (102))))
'(40 (70 (102) (52)) (28 (32) (9)))

树节点的反向节点

2 个答案: