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. Example: BSTIterator iterator = new BSTIterator(root); iterator.next(); // return 3 iterator.next(); // return 7 iterator.hasNext(); // return true iterator.next(); // return 9 iterator.hasNext(); // return true iterator.next(); // return 15 iterator.hasNext(); // return true iterator.next(); // return 20 iterator.hasNext(); // return false

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree. You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

class BSTIterator {
    private Stack<TreeNode> stack = new Stack();
    private TreeNode cur;

    public BSTIterator(TreeNode root) {
        cur = root;
    }

    /** @return the next smallest number */
    public int next() {
        while(cur != null) {
            stack.push(cur);
            cur = cur.left;
        }
        TreeNode temp = stack.pop();
        cur = temp.right;
        return temp.val;
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return(cur != null || !stack.isEmpty()); //cur为null并且stack为empty时到头了，没有下一个了
    }
}