Search range in binary search tree

Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search Tree. Find all the keys of tree in range k1 to k2. i.e. print all x such that k1<=x<=k2 and x is a key of given BST. Return all the keys in ascending order.

Example
For example, if k1 = 10 and k2 = 22, then your function should print 12, 20 and 22.

       /        
          22

  /     
      12

遍历, 没用到bst的性质:

public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
        ArrayList<Integer>  result = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        
        if (root == null) {
            return result;
        }
        
        TreeNode cur = root;
        while (! stack.isEmpty() || cur != null) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            
            cur = stack.pop();
			//中序遍历的改动点
            if (k1 <= cur.val && cur.val <= k2) {
                result.add(cur.val);
            }
            cur = cur.right;
        }
        return result;*/
    }	

递归

public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
        results = new ArrayList<Integer>();
        helper(root, k1, k2);
        return results;
    }
    
    private void helper(TreeNode root, int k1, int k2) {
        if (root == null) {
            return;
        }
        if (root.val > k1) {
            helper(root.left, k1, k2);
        }
        if (root.val >= k1 && root.val <= k2) {
            results.add(root.val);
        }
        if (root.val < k2) {
            helper(root.right, k1, k2);
        }
}