Description:

设计实现一个带有下列属性的二叉查找树的迭代器：

• 元素按照递增的顺序被访问（比如中序遍历）
• next()和hasNext()的询问操作要求均摊时间复杂度是O(1)

Explanation:

对于下列二叉查找树，使用迭代器进行中序遍历的结果为 [1, 6, 10, 11, 12]

   10
 /    \
1      11
 \       \
  6       12

Challenge:

额外空间复杂度是O(h)，其中h是这棵树的高度

Super Star：使用O(1)的额外空间复杂度

 Solution:

使用堆栈记录当前root 中序遍历的路径，将左节点从上至下推入到堆栈中。

堆栈非空，则有下一个节点。

next就是将堆栈中的节点依次进行中序遍历，返回当前节点，并将后来中序遍历的左节点推入栈中。

/**
 * 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.
    Stack<TreeNode> tree = new Stack<TreeNode>();
    
    public BSTIterator(TreeNode root) {
        // write your code here
        while(root != null){
            tree.push(root);
            root = root.left;
        }
    }

    //@return: True if there has next node, or false
    public boolean hasNext() {
        // write your code here
        return !tree.isEmpty();
    }
    
    //@return: return next node
    public TreeNode next() {
        // write your code here
        TreeNode result = tree.pop();
        TreeNode current =  result;
        
        if(current.right != null){
            current = current.right;
            while(current != null){
                tree.push(current);
                current = current.left;
            }
        }
        
        return result;
    }
}