Two Sum BSTs in Linear Time

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Medium-level problem: https://leetcode.com/problems/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

Likely the author is expecting an O(nlogn) solution, but there is a way to do in O(n) if you're willing to spare some extra space:

  • Push the items from the first tree (non-recursively) into a Hashtable
  • Go thru each item in the second tree looking for Target-Item in the Hashtable
Runs in 120ms. Code is below, cheers, ACC.


public class Solution
{
    public bool TwoSumBSTs(TreeNode root1, TreeNode root2, int target)
    {
        if (root1 == null || root2 == null) return false;
        Hashtable ht = new Hashtable();

        Queue queue = new Queue();
        queue.Enqueue(root1);
        while (queue.Count > 0)
        {
            TreeNode tn = (TreeNode)queue.Dequeue();
            if (!ht.ContainsKey(tn.val)) ht.Add(tn.val, true);

            if (tn.left != null) queue.Enqueue(tn.left);
            if (tn.right != null) queue.Enqueue(tn.right);
        }

        queue.Enqueue(root2);
        while (queue.Count > 0)
        {
            TreeNode tn = (TreeNode)queue.Dequeue();
            if (ht.ContainsKey(target - tn.val)) return true;

            if (tn.left != null) queue.Enqueue(tn.left);
            if (tn.right != null) queue.Enqueue(tn.right);
        }

        return false;
    }
}

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