leetcode------Binary Search Tree Iterator

标题：	Binary Search Tree Iterator
通过率：	28.9%
难度：	中等

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.

就是非递归中序遍历二叉树：

代码如下：

 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 public class BSTIterator {
12     TreeNode node=null;
13     Stack<TreeNode> stack=new Stack<TreeNode>();
14     public BSTIterator(TreeNode root) {
15         if(root!=null){
16             stack.push(root);
17             node=root;
18             while(node.left!=null){
19                 stack.push(node.left);
20                 node=node.left;
21             }
22         }
23     }
24 
25     /** @return whether we have a next smallest number */
26     public boolean hasNext() {
27         if(!stack.isEmpty())return true;
28         else return false;
29     }
30 
31     /** @return the next smallest number */
32     public int next() {
33         node=stack.pop();
34         int res=node.val;
35         if(node.right!=null){
36             stack.push(node.right);
37             node=node.right;
38             while(node.left!=null){
39                 stack.push(node.left);
40                 node=node.left;
41             }
42         }
43         return res;
44     }
45 }
46 
47 /**
48  * Your BSTIterator will be called like this:
49  * BSTIterator i = new BSTIterator(root);
50  * while (i.hasNext()) v[f()] = i.next();
51  */