Convert a sorted linked list to a balanced BST

The efficient solution simulates inorder construction. First count n nodes. A helper build(size) builds a balanced BST for the next size list nodes: recursively build its left subtree with size / 2 nodes, make the current list node the root, advance the shared list pointer once, then build the right subtree with the remaining nodes. Because inorder traversal visits left, root, right, consuming the sorted list in exactly that order preserves BST order and the split sizes keep height balanced. This avoids scanning for a midpoint at every recursion level. Test null, one-node, and even-length lists. Counting plus construction visits each node once, so time is O(n); recursion uses O(log n) stack space for a balanced tree.