Binary Search Tree Iterator

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

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

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

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

iterative inorder traveral 的变型。

 1 /**

 2  * Definition for binary tree

 3  * public class TreeNode {

 4  *     int val;

 5  *     TreeNode left;

 6  *     TreeNode right;

 7  *     TreeNode(int x) { val = x; }

 8  * }

 9  */

10 

11 // iterative inorder traveral change

12 public class BSTIterator {

13     LinkedList<TreeNode> stack;

14     boolean flg;

15     TreeNode tmp;

16     public BSTIterator(TreeNode root) {

17         stack = new LinkedList<TreeNode> ();

18         tmp = root;

19         flg = (root != null);

20     }

21 

22     /** @return whether we have a next smallest number */

23     public boolean hasNext() {

24         return flg && !(stack.isEmpty() && tmp == null);

25     }

26 

27     /** @return the next smallest number */

28     public int next() {

29         while(true){

30             if(tmp != null){

31                 stack.push(tmp);

32                 tmp = tmp.left;

33             }else{

34                 TreeNode result = stack.pop();

35                 tmp = result.right;

36                 return result.val;

37             }

38         }

39     }

40 }

41 

42 /**

43  * Your BSTIterator will be called like this:

44  * BSTIterator i = new BSTIterator(root);

45  * while (i.hasNext()) v[f()] = i.next();

46  */