/**

* Definition for binary tree

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode(int x) { val = x; }

* }

*/

public class BSTIterator {

private final Deque<TreeNode> stack = new ArrayDeque<>();

public BSTIterator(TreeNode root) {

advanceToNext(root);

}

/** @return whether we have a next smallest number */

public boolean hasNext() {

return !stack.isEmpty();

}

/** @return the next smallest number */

public int next() {

if (!hasNext()) {

throw new java.util.NoSuchElementException();

}

TreeNode node = stack.pop();

advanceToNext(node.right);

return node.val;

}

private void advanceToNext(TreeNode curNode) {

while (curNode != null) {

stack.push(curNode);

curNode = curNode.left;

}

}

//Another solution is using Morris traversal,

//which has the same time complexity, but the space complexity is O(1).

//https://discuss.leetcode.com/topic/8154/morris-traverse-solution

}

/**

* Your BSTIterator will be called like this:

* BSTIterator i = new BSTIterator(root);

* while (i.hasNext()) v[f()] = i.next();

*/