Question

Formatted question description: https://leetcode.ca/all/1214.html

 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:
   2     1
  / \   / \
 1   4 0   3

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

 @tag-array

Algorithm

Get two ordered list, and 2 pointers to move forward for each.

Code

Java

import java.util.ArrayList;
import java.util.List;


public class Two_Sum_BSTs {

    // Time: O(M+N)
    // Space: O(logmax(M,N)) in the best case; O(max(M,N)) in the worst case.
    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {

        // Convert to list
        List<Integer> list1 = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();
        listByInorder(root1, list1);
        listByInorder(root2, list2);

        // Two Pointers
        int lo = 0, hi = list2.size() - 1;
        while (lo < list1.size() && hi >= 0) { // notice the condition here
            int sum = list1.get(lo) + list2.get(hi);
            if (sum == target) {
                return true;
            } else if (sum < target) {
                ++lo;
            } else {
                --hi;
            }
        }
        return false;
    }

    private void listByInorder(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        listByInorder(root.left, list);
        list.add(root.val);
        listByInorder(root.right, list);
    }
}

// OJ: https://leetcode.com/problems/two-sum-bsts/
// Time: O(A + B)
// Space: O(HA + HB)
class Solution {
    void pushNodes(stack<TreeNode*> &s, TreeNode *n, bool right = false) {
        for (; n; n = right ? n->right : n->left) s.push(n);
    }
public:
    bool twoSumBSTs(TreeNode* a, TreeNode* b, int target) {
        stack<TreeNode*> sa, sb;
        pushNodes(sa, a);
        pushNodes(sb, b, true);
        while (sa.size() && sb.size()) {
            int sum = sa.top()->val + sb.top()->val;
            if (sum == target) return true;
            if (sum < target) {
                auto n = sa.top();
                sa.pop();
                pushNodes(sa, n->right);
            } else {
                auto n = sb.top();
                sb.pop();
                pushNodes(sb, n->left, true);
            }
        }
        return false;
    }
};

```
print("Todo!")
```

1214 - Two Sum BSTs

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