Oct 08, 2026 by Offer

TikTok Interview Questions 2027: Construct Binary Tree

For this TikTok interview question — build a tree from preorder and inorder — recurse root-first: preorder's first element is the root, its inorder position (via hash map) splits left and right subtrees. Pass index ranges, not slices, and state the key insight upfront: preorder gives the root, inorder gives the split.

TikTok Interview Questions: What the Tree Construction Problem Tests

Constructing a tree from traversals is commonly reported by candidates in TikTok interviews because it tests whether you truly understand what each traversal order encodes — not just that you can recite definitions. Interviewers are testing recursive decomposition (splitting the problem at the root), index arithmetic under pressure, and the hash-map optimization that takes the solution from O(n²) to O(n). Sloppy index handling is where most candidates lose points.

TikTok Interview Questions: How to Solve Step by Step

State the insight. "Preorder's first element is always the current subtree's root; that root's position in inorder divides the remaining elements into left and right subtrees." This sentence is the whole problem.
Build the inorder map. Hash map from value to index — O(1) root lookups instead of scanning. (Assume distinct values, as the problem guarantees.)
Design the recursion signature. build(preLeft, preRight, inLeft, inRight) — index ranges, no array slicing, so the solution stays O(n).
Write the base case. If preLeft > preRight, return null — empty subtree.
Build the node. root = preorder[preLeft]; find rootIdx in the map; leftSize = rootIdx - inLeft; root.left = build(preLeft+1, preLeft+leftSize, inLeft, rootIdx-1); root.right = build(preLeft+leftSize+1, preRight, rootIdx+1, inRight).
Trace a small example. Walk through preorder [3,9,20,15,7], inorder [9,3,15,20,7] showing the first split — it validates your index math live.

Example line: "Every recursive call is the same question at a smaller scale — preorder hands me the root, inorder tells me where to cut, and the map makes the cut O(1)."

Common Mistakes

Scanning inorder for the root each time. O(n²) total — the hash map is the expected optimization.
Slicing arrays. Creates O(n²) copying; index ranges keep it O(n).
Off-by-one in the preorder ranges. The leftSize computation is the fiddliest line — derive it carefully, then trace.

Keep Reading

FAQ

What are the complexities? O(n) time with the map, O(n) space for the map plus recursion stack.

Why do values need to be distinct? Otherwise the inorder position of the root is ambiguous — the problem guarantees uniqueness.

Can I build from preorder and postorder? Not uniquely — that pair is ambiguous for single-child nodes; mention this if asked as a follow-up.

Should I validate the inputs? A brief length check is nice, but don't over-engineer — focus on the construction.

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