Boundary of Binary Tree
Given a binary tree, return the values of its boundary nodes, ordered anti-clockwise, starting from the root. The boundary includes the left boundary, leaves, and right boundary.
Why Interviewers Ask This
Oracle evaluates this question to assess a candidate's mastery of tree traversal algorithms and their ability to handle complex edge cases without duplicating nodes. It specifically tests logical reasoning in defining overlapping boundaries, such as distinguishing between the leftmost path and leaf nodes, which requires precise state management during recursion.
How to Answer This Question
1. Clarify the definition: Confirm that the root is included, the left boundary excludes leaves, and the right boundary excludes leaves, ensuring no node appears twice. 2. Break down the problem: Propose solving it by collecting three distinct lists—left boundary (top-down), all leaves (in-order), and right boundary (bottom-up). 3. Define traversal logic: Explain how to traverse left children first for the left boundary, use DFS for leaves, and reverse the right boundary collection. 4. Handle edge cases: Discuss scenarios like single-node trees or skewed trees where left/right boundaries might overlap with leaves. 5. Optimize complexity: State that the solution runs in O(N) time with O(H) space for recursion stack, emphasizing efficiency suitable for Oracle's large-scale systems.
Key Points to Cover
- Explicitly define how to prevent duplicate nodes at the intersection of boundaries
- Demonstrate understanding of the specific anti-clockwise ordering requirement
- Explain the distinction between 'left boundary' nodes and 'leaf' nodes clearly
- Propose an O(N) time complexity solution using DFS traversals
- Mention handling edge cases like single-node trees or missing children
Sample Answer
To solve the Boundary of Binary Tree problem, I would first clarify the requirements to ensure we don't double-count nodes. The goal is to return values in anti-clockwise order: root, left boundary, leaves, then right boundary. My approach involves three specific traversals. First, I will collect the left boundary by moving from the root down to the leftmost leaf, adding only non-leaf nodes to avoid duplication. If a node has no left child but has a right child, I follow the right child to maintain the boundary definition. Second, I will perform a standard Depth-First Search to identify all leaf nodes, regardless of their position, adding them to our result list. This ensures every leaf is captured exactly once. Third, I will traverse the right boundary from bottom to top. Since recursive calls naturally go top-down, I will store these nodes and reverse the list before appending them to the final result. A critical detail is handling the root; if the tree has only one node, it should be returned immediately without reprocessing. For example, in a tree with a root of 1, left child 2, and right child 3, the output would be [1, 2, 3]. This method ensures O(N) time complexity and handles Oracle's strict requirement for memory efficiency and correctness in hierarchical data structures.
Common Mistakes to Avoid
- Including the root node multiple times if not carefully managed across sections
- Failing to reverse the right boundary list, resulting in clockwise instead of anti-clockwise order
- Counting leaf nodes that are also part of the left or right boundary twice
- Not handling skewed trees where the left or right boundary might consist of only right or left children respectively
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