Lowest Common Ancestor of a BST
Given a Binary Search Tree (BST) and two nodes $p$ and $q$, find their Lowest Common Ancestor (LCA). Use the BST property to avoid full tree traversal.
Why Interviewers Ask This
Apple interviewers ask this to evaluate your ability to leverage specific data structure properties rather than relying on brute-force algorithms. They want to see if you recognize that BST ordering allows for O(h) time complexity instead of O(n). This tests your logical deduction, understanding of tree traversal mechanics, and capacity to write clean, efficient code under pressure.
How to Answer This Question
1. Clarify the input: Confirm if nodes p and q exist in the tree and verify the BST property (left < root < right).
2. Define the strategy: State clearly that you will use the BST property to prune the search space, avoiding a full traversal.
3. Walk through logic: Explain the comparison process. If both values are less than the current node, move left. If both are greater, move right. If they split (one smaller, one larger), the current node is the LCA.
4. Handle edge cases: Mention scenarios where one node is an ancestor of the other or if either node matches the root.
5. Analyze complexity: Explicitly state the time complexity as O(h) where h is height, and space as O(1) for iterative or O(h) for recursive approaches.
6. Code implementation: Write clean code with meaningful variable names, adding comments that explain the decision logic at each step.
Key Points to Cover
- Explicitly state how the BST property reduces time complexity from O(n) to O(h)
- Demonstrate clear logic for handling the three cases: both left, both right, or split
- Identify the edge case where one node is the direct ancestor of the other
- Analyze space complexity differences between iterative and recursive solutions
- Use concrete numerical examples to validate the logic before coding
Sample Answer
To solve the Lowest Common Ancestor problem efficiently in a Binary Search Tree, I would leverage the inherent ordering property of BSTs. Unlike a general binary tree, we don't need to traverse the entire structure. My approach starts by comparing the values of nodes p and q with the current root node.
First, if both p and q have values smaller than the root, their common ancestor must reside in the left subtree. Therefore, I move my pointer to the left child. Conversely, if both values are larger than the root, the target lies in the right subtree, so I move right. The critical moment occurs when the values split; if one value is smaller and the other is larger than the current root, or if the root itself equals one of the nodes, then the current root is the lowest common ancestor. This is because it is the first node where the paths to p and q diverge.
For example, given a tree rooted at 6 with p=2 and q=8, since 2 < 6 and 8 > 6, 6 is immediately identified as the LCA. In a case where p=2 and q=4, both are less than 6, so we move left to node 2. Since 4 is greater than 2 but 2 is equal to p, node 2 becomes the LCA. This iterative approach ensures we find the answer in O(h) time, which is optimal for balanced trees, and uses O(1) space, demonstrating efficiency that aligns with Apple's focus on performance-critical engineering.
Common Mistakes to Avoid
- Failing to utilize BST properties and treating it like a generic binary tree traversal
- Ignoring the scenario where one node is an ancestor of the other
- Not clarifying whether the solution should be iterative or recursive regarding space usage
- Overlooking null checks or empty tree inputs in the initial setup
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