← Bytedance Interview Insights
I knew the trick going in: preorder gives you the root, inorder tells you where to split left and right subtrees.
Use the fact that the first element of preorder is the root, then find its index in inorder to split left and right subtrees. Recursively build the tree, using a hash map to achieve O(1) index lookups and O(n) overall time.
Pro tip: Mention that you can avoid slicing arrays by passing index ranges, which reduces space complexity and shows attention to performance. Also, discuss handling edge cases like empty arrays or duplicate values (if allowed) to demonstrate thoroughness.
The first element of the preorder array is the root of the current subtree.
Find the index of the root value in the inorder array; this splits the inorder into left and right subtrees.
Calculate the number of nodes in the left subtree (root index in inorder) to know how many elements belong to the left in preorder.
Recursively construct the left and right subtrees using the corresponding subarrays of preorder and inorder.
Precompute a hash map from value to index in inorder to achieve O(1) lookups, reducing time complexity to O(n).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.