The "Binary Search Tree Iterator II interview question" is an advanced version of the BST iterator. In addition to next() and hasNext(), you must also support prev() and hasPrev(). This means you need to be able to move both forward and backward through the sorted sequence of the tree's values.
Meta asks the "Binary Search Tree Iterator II coding problem" to evaluate a candidate's ability to cache and manage history. Since a standard iterator only moves forward, you need a way to "remember" where you've been without losing the efficiency of the stack-based traversal. It tests "Design interview pattern" and "List" management.
This problem combines Stack-based Traversal with a Dynamic List/Cache.
next(), store them in a list.pointer index in your history list.
next(): If the pointer is not at the end of the list, move the pointer and return the value. If it is at the end, use the stack to find the next element, add it to the list, and then move the pointer.prev(): Simply move the pointer back and return the value from the list.BST: [3, 7, 9, 15, 20]
next(): Finds 3. History: [3], Pointer: 0.next(): Finds 7. History: [3, 7], Pointer: 1.prev(): Moves pointer to 0. Returns 3.next(): Moves pointer back to 1. Returns 7.next(): Finds 9 via stack. History: [3, 7, 9], Pointer: 2.next() and prev() calls.When adding "undo" or "previous" functionality to a data structure, a combination of the original logic and a history cache is the standard design pattern. This mirrors how many real-world "Undo" features are implemented.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Binary Search Tree Iterator | Medium | Solve | |
| Increasing Order Search Tree | Easy | Solve | |
| Inorder Successor in BST II | Medium | Solve | |
| Construct Binary Search Tree from Preorder Traversal | Medium | Solve | |
| Insert into a Binary Search Tree | Medium | Solve |