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# Binary Search Tree Iterator 07/28

Design an iterator over a binary search tree with the following rules:

* Elements are visited in ascending order (i.e. an in-order traversal)
* `next()` and `hasNext()` queries run in O(*1*) time in **average**.

Have you met this question in a real interview?  Yes

#### Example

For the following binary search tree, in-order traversal by using iterator is `[1, 6, 10, 11, 12]`

```
   10
 /    \
1      11
 \       \
  6       12
```

#### Challenge

Extra memory usage O(h), h is the height of the tree.

**Super Star**: Extra memory usage O(1)空间和时间复杂度 不满足条件

```java
/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Example of iterate a tree:
 * BSTIterator iterator = new BSTIterator(root);
 * while (iterator.hasNext()) {
 *    TreeNode node = iterator.next();
 *    do something for node
 * } 
 */


public class BSTIterator {
    /*
    * @param root: The root of binary tree.
    */
    private Stack<TreeNode> stack;
    public BSTIterator(TreeNode root) {
        // do intialization if necessary
        this.stack = new Stack<>();
        
        while(root!= null){
            stack.push(root);
            root = root.left;
        }
        
    }

    /*
     * @return: True if there has next node, or false
     */
    public boolean hasNext() {
        // write your code here
        return !stack.isEmpty();
    }

    /*
     * @return: return next node
     */
    public TreeNode next() {
        // write your code here
        
        TreeNode node = stack.pop();
        if(node.right != null){
            TreeNode right = node.right;
            while(right!=null){
                stack.push(right);
                right = right.left;
            }
        }
       
        return node;
    }
}
```

Morris Traversal 非递归，不用stack，O(1 )空间复杂度