Questions: Link-Cut Trees

Question 1

What is the role of splay trees within the link-cut tree data structure?

• A. Each node in the represented forest is stored in a single global splay tree
• B. Each preferred path in the represented forest is stored as a splay tree, keyed by depth, so that path operations (find root, aggregate, update) can be performed in O(log n) amortized time by splaying and traversing the auxiliary tree
• C. Splay trees are used only for balancing the forest, not for query operations
• D. Each child pointer in the represented forest is replaced by a splay tree of that child's subtree

Answer: (B)

Link-cut trees decompose each represented tree into preferred paths — contiguous root-to-leaf chains selected based on the most recent access. Each preferred path is stored in an auxiliary splay tree where the key is the node's depth in the represented tree (so an in-order traversal of the splay tree gives the path from shallowest to deepest). Splay trees provide O(log n) amortized operations (splay, split, join) which directly translate to O(log n) amortized link-cut tree operations. The self-adjusting property of splay trees is essential: frequently accessed paths naturally rise to the top of their auxiliary trees, giving good amortized performance without explicit balancing.

Question 2

The access(v) operation in a link-cut tree makes v the root of its auxiliary splay tree and makes the path from v to the root of its represented tree the preferred path. This operation runs in O(n) worst-case time but O(log n) amortized time.

• T. True
• F. False

Answer: True

Access(v) works by: (1) splaying v within its current auxiliary tree, (2) detaching the right child of v in the auxiliary tree (cutting the preferred path below v), (3) following the path-parent pointer to the next auxiliary tree, splaying the node there, joining v's tree as the right child, and repeating until reaching the root. Each step involves a splay and a join. The worst case of a single access is O(n) — if v is at the bottom of a long non-preferred path, every step changes a preferred child. But the amortized analysis (using the same potential function as splay trees, Phi = sum log(size of subtree)) shows the amortized cost is O(log n) per access, and all other operations (link, cut, find-root) are implemented using O(1) accesses plus O(1) additional work.

Question 3

Explain how link-cut trees improve the running time of maximum flow algorithms.

Model answer: In augmenting-path max-flow algorithms (e.g., Dinic's blocking flow), each augmentation finds a path from source to sink and pushes flow along it, updating residual capacities. Without link-cut trees, finding and updating a path takes O(n) time. With link-cut trees maintaining the current tree of edges being used, the operations become: (1) find the bottleneck edge on the source-to-sink path (path minimum query, O(log n)), (2) update all edge capacities on the path (path update, O(log n)), (3) remove saturated edges and add new edges (cut and link, O(log n) each). This reduces the per-path cost from O(n) to O(log n). For Dinic's algorithm with O(n^2) blocking flow phases and O(m) paths total, link-cut trees yield O(mn log n) total time, versus O(mn^2) without them. For unit-capacity graphs, this gives O(m * sqrt(n) * log n).

The key insight is that link-cut trees turn path operations (find min, update all edges, delete saturated edge) into O(log n) operations by maintaining the augmenting tree dynamically. Each augmentation corresponds to an access, a path-minimum query, a path-update, and one or more cuts — all O(log n) amortized.

Question 4

Link-cut trees support make-tree, link, and cut operations. Which of the following correctly describes the link operation?

• A. Link(u, v) makes u a child of v in the represented forest, requiring that u is currently a root of its tree and u and v are in different trees
• B. Link(u, v) merges the splay trees of u and v without changing the represented forest
• C. Link(u, v) swaps the subtrees of u and v
• D. Link(u, v) makes v a child of u regardless of whether v is a root

Answer: (A)

Link(u, v) adds the edge (u, v) to the represented forest, making u (which must be a root of its represented tree) a child of v. The implementation: access(u) to make u the root of its auxiliary tree (it has no left child since it's the shallowest node on its preferred path after access), then set u's path-parent pointer to v. The preconditions — u must be a root and u, v must be in different trees — ensure the result is still a forest. Cut(u) removes the edge between u and its parent: access(u), then detach u's left child in the auxiliary tree (which represents the path above u).

Question 5

Link-cut trees can be used to maintain a dynamic forest under edge insertions and deletions. They cannot answer lowest-common-ancestor (LCA) queries.

• T. True
• F. False

Answer: False

Link-cut trees support LCA queries in O(log n) amortized time. To find LCA(u, v): first access(u), which makes the path from u to the root preferred. Then access(v) — as this access walks up from v toward the root, the last node where it crosses from one auxiliary tree to another (the last node on the u-to-root path that is encountered) is the LCA. Specifically, after access(u), access(v) returns the LCA as the last node that was splayed during the path-switching steps. This makes link-cut trees a powerful tool for dynamic LCA queries in addition to their primary use in path queries and dynamic connectivity.