2-Sum Binary Tree

Given a binary search tree T, where each node contains a positive integer, and an integer K, you have to find whether or not there exist two different nodes A and B such that A.value + B.value = K.

Return 1 to denote that two such nodes exist. Return 0, otherwise.

Input 1: 

T :       10
         / \
        9   20

K = 19

Return: 1

Input 2: 

T:        10
         / \
        9   20

K = 40

Return: 0

Method:

Treat BST as a sorted array, and use two pointer by using two interators

Solution:

Time: O(n)
Space: O(height of T)

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) {
 *      val = x;
 *      left=null;
 *      right=null;
 *     }
 * }
 */
public class Solution {
    public int t2Sum(TreeNode A, int B) {
        Deque<TreeNode> stack = new ArrayDeque<>();
        Deque<TreeNode> rStack = new ArrayDeque<>();
        TreeNode curr = A;
        TreeNode rCurr = A;
        while (curr != null || rCurr != null || !stack.isEmpty() || !rStack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            while (rCurr != null) {
                rStack.push(rCurr);
                rCurr = rCurr.right;
            }
            if (!stack.isEmpty() && !rStack.isEmpty()) {
                curr = stack.pop();
                rCurr = rStack.pop();
                // System.out.println("l: " + curr.val + ", r: " + rCurr.val);
                if (curr == rCurr) return 0;
                int sum = curr.val + rCurr.val;
                if (sum == B) {
                    return 1;
                } else if (sum < B) {
                    curr = curr.right;
                    rStack.push(rCurr);
                    rCurr = null;
                } else {
                    rCurr = rCurr.left;
                    stack.push(curr);
                    curr = null;
                }
            }
        }
        return 0;
    }
}