Closest Values In BST Problem

Closest Values In A BST Problem Statement

Given a non-empty binary search tree (BST) and a target value, find k values in the BST that are the closest to the target.

Example

"target": 2
"k": 2

Output:

[1, 2]

As we can see the target value is 2 and [1, 2] are the nearest values (in terms of the absolute difference between target value and node values). Besides this, the answer [2, 3] is also correct, because abs(2 – 1) = abs(2 – 3). You can return any of them. You can also return your answer in any order, that is [1, 2] or [2, 1], both are correct.

Notes

• The function accepts the root of the BST.
• Return the k nearest values to the given target value.
• The values in the BST are NOT necessarily distinct.
• If there are multiple correct answers, return any one.

Constraints:

• 1 <= number of nodes in the tree <= 100000
• 1 <= values stored in the nodes <= 10⁹
• 0 <= target value <= 10⁹
• 1 <= k <= n
• number of edges = n - 1

We have provided one solution. We will refer to the number of nodes in the given binary tree by n and number of edges by m.

• We have to find k points in the BST that are the closest to the given target value.
• We will start an in-order traversal of the given input tree.
• Besides this, we will keep a priority queue of size k to store the closest points.
• So, while traversing we keep on inserting the values in the Priority Queue until the size reaches k.
• Once the size reaches k, we check whether the current value is closer to the target value than the first value (that is the head) of the priority queue.
• If the current value is closer, we remove the head and push this value. Otherwise, we stop traversing as the further values would certainly not be closer as well because all the values after the current values will be larger than the current one.
• After the complete in-order traversal, the values remaining in the Priority Queue are the required closest points.

Time Complexity

O(n * log(k)).

We are traversing the entire input tree and at the same time inserting nodes in the priority queue if they have a possibility to be part of the final answer. So, the time needed for this is O(n * log(k)).

Auxiliary Space Used

O(n + k).

We create a Priority Queue of size k to store the closest points to the target value. Besides this, we perform an inorder tree traversal which takes O(n) stack space due to recursion. So, the total auxiliary space used is O(n + k).

Space Complexity

O(n + m + k).

Space taken by input: O(n + m).

Auxiliary space used: O(n + k).

Space used by output: O(k).

Hence, total space complexity will be O(n + m + k).

We hope that these solutions to the Closest values in a BST problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

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