Binary Tree Preorder Traversal (Iterative)
Given the root of a binary tree, return the preorder traversal of its nodes' values. Implement the solution iteratively using a stack.
Why Interviewers Ask This
Uber engineers frequently ask this to assess your mastery of non-recursive data manipulation. They want to see if you understand the underlying call stack mechanism and can manage memory efficiently without relying on implicit recursion. It tests your ability to simulate system behavior using explicit data structures, a critical skill for building high-performance, scalable backend services.
How to Answer This Question
1. Clarify requirements: Confirm input validation rules and edge cases like empty trees or single nodes. 2. Define the traversal logic: Verbally explain that preorder visits Root, then Left, then Right. 3. Propose the iterative strategy: State clearly that you will use an explicit stack to mimic the recursive call stack. 4. Walk through the algorithm: Detail pushing the root, popping it to process, and pushing right child before left child to ensure correct processing order. 5. Trace with an example: Manually trace the stack state with a small tree (e.g., root 1, left 2, right 3) to demonstrate correctness. 6. Analyze complexity: Explicitly state O(n) time and O(h) space complexity where h is height.
Key Points to Cover
- Explicitly stating the requirement to push the right child before the left child to maintain order
- Demonstrating understanding of how the stack mimics the recursive call stack mechanism
- Correctly handling edge cases such as null roots or leaf nodes
- Providing accurate Time Complexity analysis of O(n)
- Explaining Space Complexity as O(h) rather than O(n) for balanced trees
Sample Answer
To solve this iteratively, I first check if the root is null; if so, I return an empty list immediately. My approach relies on a stack to manually manage the traversal order since we cannot rely on the system's call stack. The core logic follows the Root-Left-Right pattern. I initialize the stack with the root node. While the stack is not empty, I pop the top node, add its value to my result list, and then push its children onto the stack. Crucially, I must push the right child first, followed by the left child. This reversal ensures that when I pop from the stack next, the left child is processed before the right one, maintaining the strict preorder sequence. For example, with a tree having root 1, left child 2, and right child 3, I push 1, pop 1, push 3 then 2. Next, I pop 2, process it, and since it has no children, I pop 3 and process it. This yields [1, 2, 3]. This method runs in O(n) time because every node is visited exactly once. The space complexity is O(h), where h is the tree height, representing the maximum number of nodes stored on the stack at any point. This iterative solution is robust and avoids potential stack overflow issues in deep trees, which aligns well with Uber's focus on reliability.
Common Mistakes to Avoid
- Pushing the left child before the right child, which results in incorrect Right-Left-Root ordering instead of Preorder
- Failing to handle the edge case where the input tree is empty, causing a null pointer exception
- Confusing the stack operations with queue operations, leading to Breadth-First Search logic instead of Depth-First
- Neglecting to mention space complexity or incorrectly claiming O(1) space usage
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