LeetCode 1214: Two Sum BSTs Solution
Master LeetCode problem 1214 (Two Sum BSTs), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1214. Two Sum BSTs
Problem Explanation
Explanation:
To solve this problem, we can use a recursive approach to traverse one of the BSTs while simultaneously checking if the complement of the current node's value exists in the other BST. We can utilize a HashSet to store the values of one of the BSTs and then traverse the other BST to check for the complement.
Algorithm:
- Traverse the first BST and store its values in a HashSet.
- Traverse the second BST and check if the complement of the current node's value (target - node.val) exists in the HashSet.
- If the complement exists, return true. Otherwise, continue traversing both BSTs.
- If no pair of nodes are found that sum up to the target, return false.
Time Complexity:
The time complexity of this approach is O(n1 + n2), where n1 and n2 are the number of nodes in the two BSTs.
Space Complexity:
The space complexity is O(n1) to store the values of the first BST in the HashSet.
: :
Solution Code
class Solution {
public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
Set<Integer> set = new HashSet<>();
traverseAndStore(root1, set);
return findComplement(root2, set, target);
}
private void traverseAndStore(TreeNode node, Set<Integer> set) {
if (node == null) return;
set.add(node.val);
traverseAndStore(node.left, set);
traverseAndStore(node.right, set);
}
private boolean findComplement(TreeNode node, Set<Integer> set, int target) {
if (node == null) return false;
if (set.contains(target - node.val)) return true;
return findComplement(node.left, set, target) || findComplement(node.right, set, target);
}
}Try It Yourself
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