Which statement is correct about IS-IS when it performs the Dijkstra algorithm?

The Dijkstra algorithm in IS-IS operates as follows:

Tree Database Initialization: The local router (root) is added to the tree database with a cost of 0.

Candidate Database Population: Neighbors of the root (from the LSDB) are placed into the candidate database with their associated costs.

Processing Nodes: The node with the lowest cost in the candidate database is moved to the tree database.

Neighbor Evaluation: For each neighbor of the newly added node (from the LSDB), if the neighbor is not already in the tree or candidate database, it is added to the candidate database. If it exists in the candidate with a higher cost, it is updated with the lower cost.

Termination: The algorithm stops when the candidate database is empty, ensuring all shortest paths are computed.

Analysis of Options:

A. Incorrect. The local router is placed directly into the tree database, not the candidate database.

B. Correct (with context). When a node is added to the tree database, its neighbors (existing in the LSDB) are evaluated. If these neighbors are not already in the tree or candidate database, they are added (not "moved") to the candidate database. The wording "moved" is technically inaccurate, but this option aligns closest with the process of populating the candidate database using LSDB entries during tree database processing.

C. Incorrect. Tuples (nodes) with the lowest cost are moved from the candidate database to the tree database, not from the tree to the LSDB. The LSDB remains static during SPF computation.

D. Incorrect. The algorithm stops when the candidate database is empty, not the tree database. The tree database grows as nodes are processed.