Closest Binary Search Tree Value
Given a non-empty Binary Search Tree and a target value, find the value in the BST that is closest to the target. Solve without recursion.
Why Interviewers Ask This
Spotify asks this to evaluate your ability to leverage BST properties for O(log n) efficiency rather than brute force. They specifically test if you can optimize space complexity by replacing recursion with an iterative approach, demonstrating mastery of tree traversal and stack management under constraints.
How to Answer This Question
1. Clarify constraints: Confirm the target is a float and the tree is non-empty. 2. Define the iterative strategy: Explain that instead of recursion, you will use a loop to traverse from the root. 3. Establish logic: State that at each node, compare the current value to the target; move left if the node is greater, right otherwise, updating the closest candidate whenever a smaller difference is found. 4. Handle edge cases: Briefly mention handling null checks or single-node trees. 5. Complexity analysis: Conclude by explicitly stating the time complexity is O(h) where h is height, and space is O(1) due to the lack of call stack usage.
Key Points to Cover
- Explicitly leveraging BST ordering to prune search space
- Demonstrating O(1) space complexity via iteration
- Correctly handling floating-point comparison logic
- Avoiding recursion depth issues for deep trees
- Clear explanation of why one branch is discarded
Sample Answer
To solve the Closest Binary Search Tree Value problem iteratively, I first acknowledge that while recursion is natural for trees, the constraint requires an iterative solution to save stack space. I would start at the root and initialize a variable, say 'closest', to the root's value. As I iterate through the tree, I calculate the absolute difference between the current node's value and the target. If this difference is smaller than the difference for 'closest', I update 'closest'. The key optimization lies in using the BST property: if the current node's value is greater than the target, the closest value must be in the left subtree (or the current node), so I move left. Conversely, if it is less, I move right. This allows me to discard half the tree at every step, achieving O(log n) average time complexity. Unlike a recursive solution which uses O(h) stack space, this iterative method maintains O(1) space complexity. For example, if the target is 6.0 and we are at a node with value 8, we know the answer cannot be in the right subtree because all values there are larger than 8, moving further away from 6. By strictly following this path until reaching a leaf, we guarantee finding the global minimum difference without ever needing to backtrack or store state on a stack.
Common Mistakes to Avoid
- Failing to update the closest value when traversing down the wrong path initially
- Using recursion despite the explicit constraint against it
- Comparing integers only and ignoring floating-point precision issues
- Not explaining how the BST property justifies moving left or right
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