Insert Node into a Tree - Racket

与世无争的帅哥 提交于 2019-12-25 16:46:49

问题


I am trying to add a new node to the tree. The following are my definitions and function type:

(define-struct (Some T)
  ([value : T]))

(define-type (Option T)
  (U 'None (Some T)))

(define-type BST (U 'E Nd))

(define-struct Nd
  ([root : Integer]
   [lsub : BST]
   [rsub : BST]))

(: insert : Integer BST -> BST)
;; insert an item into a tree
;; note: do not insert duplicate items
(define (insert n x)
  (match x
    ('E 'E)
    ((Nd ro ls rs)
     (cond
       ((= (size x) 1) (Nd ro (Nd n 'E 'E) 'E))
       (else
        (Nd ro ls rs))))))

Insert is the insert that will insert the node into the tree.

The following is the command that I will give:

(insert 10 (Nd 1 (Nd 2 (Nd 4 'E 'E) (Nd 5 'E 'E)) (Nd 3 (Nd 6 'E 'E) (Nd 7 'E 'E))))

And it should insert ten into the tree. However, I am learning independently at home and I have NO idea what to do. Please help. Thank you so much!


回答1:


You're missing the recursion, and your base case is wrong.

Inserting in an empty tree creates a tree with one node.

Inserting in a non-empty BST has three cases:

  • If the item is the same as in this node, return the tree unchanged
  • If the item is smaller than this node, insert in the left subtree
  • Otherwise, insert in the right subtree

Something like

(define (insert n x)
  (match x
    ('E (Nd n 'E 'E))
    ((Nd ro ls rs)
     (cond
      ((= n ro) x)
      ((< n ro) (Nd ro (insert n ls) rs))
      (else     (Nd ro ls (insert n rs)))))))

The tree you're aiming to insert in isn't a BST though, so this won't work.

Your tree has the following structure:

   1
  /\
 2  3
 /\ /\
4 5 6 7

A search tree with those elements would look like this:

   4
  /\
 2  6
 /\ /\
1 3 5 7

which is

(Nd 4 (Nd 2 (Nd 1 'E 'E) (Nd 3 'E 'E)) (Nd 6 (Nd 5 'E 'E) (Nd 7 'E 'E)))


来源:https://stackoverflow.com/questions/40519825/insert-node-into-a-tree-racket

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