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Problem solution · SQL

CEO Subordinate Hierarchy

CEO Subordinate Hierarchy: a SQL solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
39 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For CEO Subordinate Hierarchy, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 39 lines of SQL from the credited upstream file 3236.sql.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCEO Subordinate Hierarchy · SQLSQL
Use this to learn the idea, then write your own version.
WITH RECURSIVE  EmployeeHierarchy AS (    -- Base case: direct reports to CEO    SELECT      employee_id,      employee_name,      salary,      1 AS hierarchy_level    FROM Employees    WHERE manager_id = (        SELECT employee_id        FROM Employees        WHERE manager_id IS NULL      )    UNION ALL    -- Recursive case: reports of reports    SELECT      Employees.employee_id,      Employees.employee_name,      Employees.salary,      EmployeeHierarchy.hierarchy_level + 1    FROM Employees    INNER JOIN EmployeeHierarchy      ON (Employees.manager_id = EmployeeHierarchy.employee_id)  ),  Ceo AS (    SELECT salary    FROM Employees    WHERE manager_id IS NULL  )SELECT  EmployeeHierarchy.employee_id AS subordinate_id,  EmployeeHierarchy.employee_name AS subordinate_name,  EmployeeHierarchy.hierarchy_level,  EmployeeHierarchy.salary - Ceo.salary AS salary_differenceFROM EmployeeHierarchyCROSS JOIN CeoORDER BY hierarchy_level, employee_id;

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