Reconstruct a binary tree from its preorder and inorder traversals.

Preorder names the root; inorder splits the rest into left and right subtrees. Doing it naively is O(n squared); the strong answer uses a value-to-index map and a moving preorder pointer for O(n). Here is the answer and why both orders are required.

Updated Aug 2026 · Grounded in real Applied AI Engineer interview loops and written to a senior-engineer editorial bar.