BST Iterator

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

The first call to next() will return the smallest number in BST. Calling next() again will return the next smallest number in the BST, and so on.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
Try to optimize the additional space complexity apart from the amortized time complexity.

Solution:

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

public class Solution {
    private Deque<TreeNode> stack;
    private TreeNode curr;
    
    public Solution(TreeNode root) {
        stack = new ArrayDeque<>();
        curr = root;
    }

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

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

/**
 * Your Solution will be called like this:
 * Solution i = new Solution(root);
 * while (i.hasNext()) System.out.print(i.next());
 */

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class BSTIterator {
    
    private Deque<TreeNode> stack;
    private TreeNode node;

    public BSTIterator(TreeNode root) {
        stack = new ArrayDeque<TreeNode>();
        node = root;
    }
    
    /** @return the next smallest number */
    public int next() {
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
        TreeNode next = stack.pop();
        node = next.right;
        return next.val;
    }
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty() || node != null;
    }
    
//     private Stack<TreeNode> stack = new Stack<TreeNode>();
    
//     public BSTIterator(TreeNode root) {
//         pushAll(root);
//     }

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

//     /** @return the next smallest number */
//     public int next() {
//         TreeNode tmpNode = stack.pop();
//         pushAll(tmpNode.right);
//         return tmpNode.val;
//     }
    
//     private void pushAll(TreeNode node) {
//         for (; node != null; stack.push(node), node = node.left);
//     }
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator obj = new BSTIterator(root);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */