Question

我在使用二叉搜索树时遇到问题并将其转换为列表。

(define-struct node (key val left right))
;; A binary search tree (bst) is either
;; empty, or
;; a structure (make-node k v l r), where
;; k is a number (the key),
;; v is a string (the value),
;; l is a bst, where every key in l is less than k, and
;; r is a bst, where every key in r is greater than k.

有人可以给我一些关于如何处理这个问题的提示吗？

创建一个使用二叉搜索树的函数bst，并返回二叉搜索树节点的值字段中所有字符串的列表，并且列表必须根据二叉搜索树中的键值按降序排列。

;;Examples: (bst (make-node 4 "James" (make-node 2 "Kien" empty empty)
;;(make-node 5 "Jack" empty (make-node 11 "Cole" empty empty)))) => (list "Cole" "Jack" "James" "Kien")

谢谢！

Answer 1

基本上，您必须使用反向in-orden traversal（右子树/值/左子树）访问树中的所有节点，同时创建包含答案的列表。这些方面的东西：

(define (tree->list tree)
  (if (empty? tree)
      empty
      (append (tree->list (node-right tree))
              (cons (node-val tree)
                    (tree->list (node-left tree))))))

按预期工作：

(define bst
  (make-node 4 "James"
             (make-node 2 "Kien" empty empty)
             (make-node 5 "Jack" empty
                        (make-node 11 "Cole" empty empty))))

(tree->list bst)
=> (list "Cole" "Jack" "James" "Kien")

Answer 2

(define (tree->list tree)
  (if(leaf? tree)
     (list (node tree))
     (cond((right-branch-only? tree)(append (list(node tree))
                                            (tree->list (right-branch tree))))
          ((left-branch-only? tree)(append (list(node tree))
                                           (tree->list (left-branch tree))))
          (else(append (list (node tree))
                       (append (tree->list (left-branch tree))
                               (tree->list (right-branch tree))))))))

初学者方案：将二进制搜索树转换为列表

2 个答案: