Two Sum IV In BST: Solutions & Guide

by Viktoria Ivanova 37 views

Hey guys! Today, we're diving deep into a fascinating problem: finding two numbers in a Binary Search Tree (BST) that add up to a specific target value. This is a classic problem with a few clever solutions, and we're going to explore them all. So, buckle up and let's get started!

Understanding the Problem: Two Sum IV - Input is a BST

Before we jump into the code, let's make sure we fully grasp the problem. The prompt, often titled "Two Sum IV - Input is a BST", gives us the root of a Binary Search Tree and a target integer, k. Our mission, should we choose to accept it, is to determine if there exist two distinct nodes in the BST whose values sum up to k. In simpler terms, we need to find two numbers within the tree that, when added together, equal our target value.

Now, the fact that we're dealing with a Binary Search Tree is crucial. BSTs have a special property: the value of every node in the left subtree is less than the node's value, and the value of every node in the right subtree is greater. This ordering allows us to use efficient search algorithms. This characteristic, guys, is what we'll exploit to solve this problem efficiently.

To illustrate, imagine a BST containing the values [5, 3, 6, 2, 4, null, 7], and our target k is 9. We need to figure out if there are two nodes in this tree that add up to 9. In this case, the nodes with values 2 and 7 fit the bill, so our answer would be "true." If, however, k were 20, no such pair exists, and we'd return "false."

It's essential to understand the constraints as well. Typically, these problems come with limitations on the size of the tree (number of nodes) and the range of values the nodes can hold. Knowing these constraints helps us choose the most appropriate algorithm. For instance, if the tree is very large, we'll want to avoid solutions with high time complexity.

Approach 1: Inorder Traversal and Two-Pointer Technique

Our first approach leverages the inherent ordering of a BST. Remember how the inorder traversal of a BST gives us the nodes in sorted order? This is the key, guys! We can perform an inorder traversal, store the node values in a sorted list, and then apply the classic two-pointer technique to find our pair.

Detailed Steps

  1. Inorder Traversal: We begin by performing an inorder traversal of the BST. This involves recursively visiting the left subtree, then the current node, and finally the right subtree. As we visit each node, we append its value to a list (or array). This process ensures that our list will be sorted in ascending order.
  2. Two-Pointer Initialization: Once we have our sorted list, we initialize two pointers. Let's call them left and right. The left pointer starts at the beginning of the list (index 0), and the right pointer starts at the end of the list (index n-1, where n is the list's length).
  3. Iteration and Comparison: Now, the magic happens! We enter a while loop that continues as long as left is less than right. Inside the loop, we do the following:
    • Calculate the sum of the values at the left and right pointers: sum = list[left] + list[right].
    • Compare sum with our target k:
      • If sum equals k, we've found our pair! We return true.
      • If sum is less than k, it means we need a larger sum. To achieve this, we increment the left pointer to move to a larger value in the list.
      • If sum is greater than k, we need a smaller sum. We decrement the right pointer to move to a smaller value.
  4. No Pair Found: If the while loop completes without finding a pair (i.e., left becomes greater than or equal to right), it means no such pair exists in the tree. We return false.

Why This Works

The beauty of this approach lies in the sorted nature of the list and the efficiency of the two-pointer technique. Because the list is sorted, we can systematically move our pointers inward, knowing whether we need to increase or decrease the sum. This avoids unnecessary comparisons and gives us a time complexity of O(n), where n is the number of nodes in the tree.

Time and Space Complexity

  • Time Complexity: O(n), where n is the number of nodes in the BST. This is because the inorder traversal takes O(n) time, and the two-pointer technique also takes O(n) time in the worst case.
  • Space Complexity: O(n), as we need to store the node values in a list.

Approach 2: Using a Hash Set

Our second approach takes a different route, employing a hash set to optimize our search. This method allows us to check for the existence of a complementary value in O(1) time on average, making it a very efficient solution.

Detailed Steps

  1. Initialize a Hash Set: We start by creating an empty hash set (also known as a hash table). This set will store the values we've encountered as we traverse the tree. The fast lookup capability of hash sets is the key to this approach.
  2. Recursive Traversal: We perform a traversal of the BST. This can be either an inorder, preorder, or postorder traversal – the order doesn't matter for this method. For each node we visit, we do the following:
    • Calculate the complement: complement = k - node.val. This is the value that, when added to the current node's value, would equal our target k.
    • Check for complement in the hash set: We use the hash set's contains (or equivalent) operation to check if the complement is already present in the set. If it is, we've found our pair, and we return true.
    • Add the current node's value to the hash set: If the complement is not found, we add the current node's value (node.val) to the hash set. This way, we can check for it later when we visit other nodes.
  3. No Pair Found: If we traverse the entire tree without finding a pair (i.e., the true condition is never met), we return false.

Why This Works

The magic of this approach lies in the hash set's ability to provide near-constant-time lookups. For each node, we can quickly determine if its complement exists in the tree by checking the hash set. This drastically reduces the time complexity compared to a naive approach that would involve comparing every pair of nodes.

Time and Space Complexity

  • Time Complexity: O(n) on average, where n is the number of nodes in the BST. We visit each node once, and the hash set operations (contains and add) take O(1) time on average.
  • Space Complexity: O(n) in the worst case, as we might need to store all the node values in the hash set if no pair is found until the very end of the traversal.

Approach 3: Recursive Search with Two Trees (Less Efficient)

While the previous two approaches are the most efficient, let's explore a third approach that, while less optimal, showcases a different way of thinking about the problem. This method involves recursively searching for the complement of each node's value in the BST.

Detailed Steps

  1. Outer Traversal: We start with a traversal of the BST. For each node we visit (let's call it node1), we perform the following:
    • Calculate the complement: complement = k - node1.val
    • Inner Search: We call a helper function (or perform a separate recursive search) to look for the complement within the BST. This inner search starts from the root of the BST.
    • Handle duplicates: The inner search function must avoid considering the node that was used to derive the complement. Consider a scenario where k = 10 and the current node's value (node1.val) is 5. The complement is also 5. We should not consider node1 itself as a valid complement, because the problem requires distinct nodes.
    • If the inner search finds the complement, we've found our pair and return true.
  2. Inner Search Function: The inner search function is a standard BST search. It takes the root of the BST and the value to search for (complement) as input. It recursively does the following:
    • If the root is null, the value is not found, so return false.
    • If the root's value equals the complement, and the root is not the original node (node1), we've found our match. Return true.
    • If the complement is less than the root's value, recursively search in the left subtree.
    • If the complement is greater than the root's value, recursively search in the right subtree.
  3. No Pair Found: If the outer traversal completes without finding a pair, we return false.

Why This Works (and Why It's Less Efficient)

This approach essentially tries every node in the tree and searches for its complement. While it will eventually find a pair if one exists, the repeated searches make it less efficient than the previous methods. The inner search effectively re-traverses parts of the tree for each node in the outer traversal.

Time and Space Complexity

  • Time Complexity: O(n log n) on average, where n is the number of nodes in the BST. The outer traversal takes O(n) time, and the inner search (BST search) takes O(log n) time on average for a balanced BST. In the worst case (a skewed BST), the time complexity can be O(n^2).
  • Space Complexity: O(h), where h is the height of the BST. This is due to the recursive call stack. In the worst case (a skewed BST), h can be n, leading to O(n) space complexity. For a balanced BST, h is log n, resulting in O(log n) space complexity.

Which Approach Should You Use?

Okay, guys, so we've explored three different ways to tackle the Two Sum IV problem. But which one should you actually use in a real-world scenario or during an interview?

  • Approach 1 (Inorder Traversal and Two-Pointer): This is a solid choice with a clear O(n) time complexity. It's easy to understand and implement. The only downside is the O(n) space complexity due to the extra list we create to store the inorder traversal.
  • Approach 2 (Hash Set): This is generally the preferred approach due to its excellent average time complexity of O(n) and relatively straightforward implementation. The space complexity is also O(n) in the worst case, but the speed advantage often outweighs this.
  • Approach 3 (Recursive Search with Two Trees): This approach is the least efficient due to its O(n log n) or O(n^2) time complexity. It's a good exercise in understanding recursion and BST properties, but not recommended for practical use unless space complexity is a major constraint and a balanced BST is guaranteed.

In most cases, I'd recommend going with Approach 2 (Hash Set) for its balance of speed and simplicity. But knowing all three approaches gives you flexibility and demonstrates a deeper understanding of the problem.

Conclusion

So there you have it, guys! A comprehensive look at solving the Two Sum IV problem in a Binary Search Tree. We've covered three different approaches, analyzed their time and space complexities, and discussed when to use each one.

Remember, the key to mastering these problems is not just memorizing solutions, but understanding the underlying principles and trade-offs. Keep practicing, keep exploring, and you'll become a BST ninja in no time!