Find a pair with given sum in a Balanced BST

Given a Balanced Binary Search Tree and a target sum, write a function that returns true if there is a pair with sum equals to target sum, otherwise return false. Expected time complexity is O(n) and only O(Logn) extra space can be used. Any modification to Binary Search Tree is not allowed. Note that height of a Balanced BST is always O(Logn).

http://www.geeksforgeeks.org/find-a-pair-with-given-sum-in-bst/

private boolean twoSumBST(TreeNode root, int sum) {
    Stack<TreeNode> stk1 = new Stack<TreeNode>();
    TreeNode cur1 = root;
    Stack<TreeNode> stk2 = new Stack<TreeNode>();
    TreeNode cur2 = root;
    boolean nextV1 = true;
    boolean nextV2 = true;
int val1 = 0;
int val2 = 0;
    while (true) {
    if (nextV1) {
    while (true) {
    if (cur1 != null) {
    stk1.push(cur1);
    cur1 = cur1.left;
    } else {
    cur1 = stk1.pop();
    val1 = cur1.val;
    cur1 = cur1.right;
    break;
    }
    }
    }
   
    if (nextV2) {
    while (true) {
    if (cur2 != null) {
    stk2.push(cur2);
    cur2 = cur2.right;
    } else {
    cur2 = stk2.pop();
    val2 = cur2.val;
    cur2 = cur2.left;
    break;
    }
    }
    }
    System.out.println(val1 + "+" + val2);
    if (val1 >= val2) {
    break;
    }
    if (val1 + val2 == sum) {
    return true;
    } else if (val1 + val2 < sum) {
    nextV1 = true;
    nextV2 = false;
    } else if (val1 + val2 > sum) {
    nextV1 = false;
    nextV2 = true;
    }
    }
    return false;
    }