Binary Search Tree Iterator Solution

Question

Implement an iterator over a binary search tree (BST). Your iterator is initialized with the root node of a BST.

Calling would return the next smallest number in the next() BST.

Note: next() hasNext() and should run in average O (1) time and uses O (h) memory, where H is the height of the Tree.

Solution

When a problem relates to BST and sequence, we should think about in-order traversal of BST.

We find that the original codes of Binary Tree inorder traversal can is modified to fit following codes.

 while (It.hasnext ()) {    System.out.println (It.next ());        }

Note that before print, we need to the push nodes to stack first.

Inorder traversal time complexity is O (n), so for each next () step, average Time is O (1).

And the stack costs O (h) because we either goes down or pops a node and then goes down.

1 /**2 * Definition for binary tree3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  One  Public classBstiterator { A     PrivateStack<treenode>Stack; -     PrivateTreeNode Current; -  the      Publicbstiterator (TreeNode root) { -stack =NewStack<treenode>(); -Current =Root; -          while(Current! =NULL) { + Stack.push (current); -Current =Current.left; +         } A          at     } -  -     /** @returnwhether we have a next smallest number*/ -      Public BooleanHasnext () { -         return(! Stack.empty () | | | current! =NULL); -     } in  -     /** @returnThe next smallest number*/ to      Public intNext () { +          while(Current! =NULL) { - Stack.push (current); theCurrent =Current.left; *         } $TreeNode tmp =Stack.pop ();Panax Notoginseng         intresult =Tmp.val; -Current =Tmp.right; the         returnresult; +     } A } the  + /** - * Your Bstiterator 'll be a called like this: $ * Bstiterator i = new Bstiterator (root); $ * while (I.hasnext ()) v[f ()] = I.next (); -  */

Binary Search Tree Iterator Solution