Write a self join query to get the manager's name for each employee.

A self join treats one table as two logical tables so you can relate rows to other rows in the same table. For an employee list where each row stores its own manager's id, you join employees to itself: once as the employee, once as the manager.

SELECT e.name  AS employee_name,
       m.name  AS manager_name
FROM   employees e
LEFT   JOIN employees m
       ON e.manager_id = m.id;

Why LEFT JOIN and not INNER JOIN

The alias is what makes this work: e and m are two independent cursors over the same physical table, so the optimizer treats them as separate inputs. Use a LEFT JOIN because the top of the hierarchy has a NULL manager_id. An INNER JOIN silently drops the CEO, and quietly losing rows is the single most common mistake on this question. If you want a friendlier output, wrap the result: COALESCE(m.name, 'No manager').

Going deeper than one level

A self join only climbs one level. To walk an arbitrary-depth org chart you need a recursive CTE:

WITH RECURSIVE chain AS (
  SELECT id, name, manager_id, 1 AS depth
  FROM   employees
  WHERE  manager_id IS NULL
  UNION ALL
  SELECT e.id, e.name, e.manager_id, c.depth + 1
  FROM   employees e
  JOIN   chain c ON e.manager_id = c.id
)
SELECT * FROM chain;

Production considerations

Index manager_id, since it is the join key and is usually unindexed on a table that was designed around id. Real org data also contains cycles from bad imports, and a recursive CTE on a cycle runs until it exhausts memory, so guard it with a depth cap.

In the interview, also mention that the same self-join pattern solves "find employees earning more than their manager" by adding WHERE e.salary > m.salary.