Binary Search Tree Iterator

題目：寫一個BST遍歷器
1. 初始化遍歷器
2. next() will return the next smallest number in the BST
3. next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree

解題思路：
因爲BST的最小值永遠在最左邊，如果沒有限制的話，只要用一個隊列來存中序遍歷的結果，即可；
可是這裏只能用O(h)的空間；h爲高度，所以，可以用一個棧來存根節點的所有左孩子，棧頂永遠存的是最小值，當next()
的時候，如果節點有右孩子，則以它爲根節點，所有的左孩子入棧；

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {

    LinkedList<TreeNode> stack = new LinkedList<TreeNode> ();

    public BSTIterator(TreeNode root) {
        addToStack(root);
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode node = stack.removeLast();
        addToStack(node.right);
        return node.val;
    }

    private void addToStack(TreeNode root) {
        while(root != null) {
            stack.add(root);
            root = root.left;
        }
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */