Closest Binary Search Tree Value II

Problem Statement

The task is rooted in a common problem involving binary search trees (BSTs), where one needs to find the k closest values to a given target number within the tree. Working with BSTs is beneficial due to their properties of ordered data, which can potentially simplify searching and sorting operations. Given a target value and an integer k, the goal is to accurately identify the k nearest numbers to the target from the numbers present in the BST. The output does not necessitate a specific order, hence any sequence containing the correct values is valid. The problem assures that there is a unique set of k values closest to the target, which simplifies the task by avoiding ambiguity in determining which numbers are closest when distances are identical.

Examples

Example 1

Input:

root = [4,2,5,1,3], target = 3.714286, k = 2

Output:

[4,3]

Example 2

Input:

root = [1], target = 0.000000, k = 1

Output:

[1]

Constraints

• The number of nodes in the tree is n.
• 1 <= k <= n <= 104.
• 0 <= Node.val <= 109
• -109 <= target <= 109

Approach and Intuition

1. BST Traversal:

   • A straightforward approach could involve an in-order traversal of the BST, which inherently produces a sorted list of elements. This sorted list can then be used to find the k elements nearest to the target.

2. Optimized Search Using a Heap:

   • Rather than sorting the complete list of elements, one can maintain a heap (or priority queue) of size k to store the closest elements encountered during a tree traversal. If we use a max-heap, we can compare the absolute difference of the current node value with the target. If this difference is less than the maximum difference in the heap, we replace the maximum element in the heap with the current element.
   • This approach ensures that at any point, the heap contains the k closest elements, and since we are employing a max-heap, the element with the largest difference (least close to the target) can be efficiently removed if needed.

3. Balanced Traversal Algorithm:

   • To further improve, particularly if k is much smaller than n, a more directed search using the properties of BST can be deployed. Starting from the root node, one can decide to move left or right depending on how the current node's value compares to the target. This reduces the number of nodes to consider since half of the tree can potentially be disregarded at each step based on whether the target value is less than or greater than the node’s value.

This problem lays the foundation for understanding some fundamental operations with binary trees, such as traversal techniques and dynamic data storage using heaps, while also requiring efficient decision-making based on the tree's properties and the problem's constraints.

Solutions

class Solution {
    public List<Integer> findClosestElements(TreeNode root, double target, int k) {
        Deque<Integer> result = new LinkedList<>();
        traverseInOrder(root, result, k, target);
        return new ArrayList<>(result);
    }
    
    public void traverseInOrder(TreeNode current, Deque<Integer> result, int k, double target) {
        if (current == null) {
            return;
        }
    
        traverseInOrder(current.left, result, k, target);
        result.add(current.val);
        if (result.size() > k) {
            if (Math.abs(target - result.peekFirst()) <= Math.abs(target - result.peekLast())) {
                result.removeLast();
                return;
            } else {
                result.removeFirst();
            }
        }
    
        traverseInOrder(current.right, result, k, target);
    }
}