← Bytedance Interview Insights
I jumped straight to checking left child less than root and right child greater than root, which is the classic wrong answer.
Use a recursive approach that passes down the allowed range (min, max) for each node. At each node, check if its value is within the range, then recursively validate the left subtree with an updated max and the right subtree with an updated min. This ensures all nodes satisfy the BST property with respect to all ancestors, not just their immediate parent.
Pro tip: Mention that an inorder traversal should yield a strictly increasing sequence, and you can solve it iteratively with O(1) space using Morris traversal if asked for optimization. Also, clarify whether duplicate values are allowed, as this affects the strictness of the inequality.
Ask if the tree can contain duplicate values and whether they should be considered valid. Confirm the definition of a BST (left < root < right, or left <= root < right).
Decide between recursive range-checking or inorder traversal. For interviews, recursive range-checking is straightforward and easy to explain; inorder traversal is also valid and can be done iteratively.
Write a helper function that takes a node and a (min, max) range. If the node is null, return true. If node.val <= min or node.val >= max, return false. Recurse left with (min, node.val) and right with (node.val, max).
Consider empty tree (return true), single node (return true), and trees with extreme values (use null or infinity for initial bounds). Also, test with a tree that is not a BST but satisfies local ordering.
State that the time complexity is O(n) since each node is visited once, and space complexity is O(h) for recursion stack, where h is the height of the tree.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.