Two Sum BSTs

Given two binary search trees, return True if and only if there is a node in the first tree and a node in the second tree whose values sum up to a given integer target.

Example 1:

Input: root1 = [2,1,4], root2 = [1,0,3], target = 5
Output: true
Explanation: 2 and 3 sum up to 5.

Example 2:

Input: root1 = [0,-10,10], root2 = [5,1,7,0,2], target = 18
Output: false

Constraints:

Each tree has at most 5000 nodes.
-10^9 <= target, node.val <= 10^9

Solution:

/**
 * 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 Solution {
    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        List<Integer> list = new ArrayList();
        in(root1, list);
        for (int val : list) {
            if (bs(root2, target - val)) {
                return true;
            }
        }
        return false;
    }
    
    private void in(TreeNode root, List<Integer> list) {
        if (root == null) return;
        in(root.left, list);
        list.add(root.val);
        in(root.right, list);
    }
    
    private boolean bs(TreeNode root, int target) {
        if (root == null) return false;
        if (root.val == target) {
            return true;
        } else if (root.val < target) {
            return bs(root.right, target);
        } else {
            return bs(root.left, target);
        }
    }
}