NVIDIA·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026

Summary

NVIDIA software engineer interview with a tree reconstruction problem. Nothing too wild but it's the kind of question that looks straightforward until you're actually coding it under pressure.

Questions Asked (1)

Given the inorder and postorder traversal arrays of a binary tree, reconstruct the original tree and return its root.

Algorithms & Data Structures

Author's notes

The core insight is that the last element of postorder is always the root, then you find that value in inorder to split left and right subtrees.

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AI HintsAI Generated

Suggested Approach

Explain that the last element of the postorder array is the root, and its position in the inorder array splits the tree into left and right subtrees. Recursively apply this logic to reconstruct the tree, using a hash map for O(1) index lookups to achieve O(n) time complexity.

Pro tip: Mention that you can avoid copying subarrays by passing indices, which reduces space complexity and shows attention to performance—a key trait at NVIDIA where efficiency matters.

1. Identify the root

The last element of the postorder traversal is the root of the current subtree.

2. Split inorder

Find the root's index in the inorder array; elements to the left form the left subtree, and elements to the right form the right subtree.

3. Recursively build subtrees

Use the corresponding segments of the postorder array to recursively construct the left and right subtrees.

4. Optimize with hash map

Precompute a hash map from value to index in the inorder array to achieve O(1) lookups, reducing overall time complexity to O(n).

5. Handle edge cases

Check for empty arrays, single-node trees, and skewed trees; ensure base cases return null appropriately.

Key Points to Mention

Time and space complexity: O(n) time with hash map, O(n) space for recursion stack and hash map.
Recursive approach with index boundaries to avoid array copying.
Handling duplicate values (if allowed) by using a hash map that stores the last occurrence or by assuming uniqueness.
Base case: when the inorder segment is empty, return null.
Comparison with preorder+inorder reconstruction and why postorder requires processing from the end.
Potential iterative solution using a stack for further optimization.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.