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.

Example**For the following binary search tree, in-order traversal by using iterator is [1, 6, 10, 11, 12]
············10
···········/ ···
·········1·····11
···········\······ \
············6 ·····12

題目要求next()返回下一個最小的數，其實不難看出是中序遍歷的順序，二叉樹一個性質就是中序遍歷是從小到大的遞增數列，其實這題就是寫一個中序遍歷的迭代器。
遍歷二叉樹有遞歸和迭代的方法，這題用迭代，當然就要用到stack這種后進先出的數據結構。
代碼初始化BSTIterator（）的時候就把root和其最左的那條路徑上的nodes都放進stack。
next()方法中，我們先取出當前的準備返回的node， 然后check他的右子樹是否為空，如果為空當然不用做什么，如果不為空，我們就先把它push到stack里面去，然后遍歷它的左子樹放進stack。

/**
 * 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> stack = new Stack<>();
    public BSTIterator(TreeNode root) {
        // write your code here
        while(root != null) {       //firs push root and all the left most line nodes to stack
            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) {        //if the poped out node has right subtree, we need push the nodes in stack
            
            TreeNode nextPush = node.right;
            stack.push(nextPush);                       //push the right node
            while(nextPush.left != null ) {            // push the left most line of nodes of the right node
                stack.push(nextPush.left);
                nextPush = nextPush.left;
            }
        }
        
        return node;
    }
}