Recursive CTE for graph traversal in PostgreSQL

Introduction

Recursive CTEs help you explore connected data step-by-step, like finding all friends of a friend in a network.

Finding all connected nodes in a network, like friends in a social media graph.

Tracing paths in a family tree to find ancestors or descendants.

Exploring dependencies between tasks or projects.

Finding all reachable locations from a starting point on a map.

Analyzing organizational charts to find all subordinates under a manager.

Syntax

PostgreSQL

WITH RECURSIVE cte_name(column_list) AS (
    -- Anchor member: starting point of recursion
    SELECT initial_columns
    FROM table_name
    WHERE condition
  
    UNION ALL
  
    -- Recursive member: references cte_name to find connected rows
    SELECT next_columns
    FROM table_name
    JOIN cte_name ON join_condition
    WHERE recursive_condition
)
SELECT * FROM cte_name;

The anchor member is the base query that starts the recursion.

The recursive member references the CTE itself to find connected rows.

Examples

Finds all friends and friends of friends starting from user 1, avoiding cycles back to user 1.

PostgreSQL

WITH RECURSIVE friends_cte(friend_id) AS (
  SELECT friend_id FROM friendships WHERE user_id = 1
  UNION ALL
  SELECT f.friend_id FROM friendships f
  JOIN friends_cte fc ON f.user_id = fc.friend_id
  WHERE f.friend_id != 1
)
SELECT * FROM friends_cte;

Finds all ancestors of person with id 5 by recursively joining on parent_id.

PostgreSQL

WITH RECURSIVE ancestors_cte(id, parent_id) AS (
  SELECT id, parent_id FROM family WHERE id = 5
  UNION ALL
  SELECT f.id, f.parent_id FROM family f
  JOIN ancestors_cte a ON f.parent_id = a.id
)
SELECT * FROM ancestors_cte;

Sample Program

This query finds all nodes you can reach starting from node 1 by following edges step-by-step.

PostgreSQL

CREATE TABLE edges (
  from_node INT,
  to_node INT
);

INSERT INTO edges VALUES
  (1, 2),
  (1, 3),
  (2, 4),
  (3, 5),
  (5, 6);

-- Find all nodes reachable from node 1
WITH RECURSIVE reachable_nodes(node) AS (
  -- Start from node 1
  SELECT 1 AS node
  UNION
  -- Find nodes connected to already found nodes
  SELECT e.to_node FROM edges e
  JOIN reachable_nodes r ON e.from_node = r.node
)
SELECT DISTINCT node FROM reachable_nodes ORDER BY node;

Important Notes

Time complexity depends on the number of edges and nodes; it can be expensive for large graphs.

Space complexity grows with the number of reachable nodes stored during recursion.

Common mistake: forgetting to include a stopping condition or DISTINCT, which can cause infinite loops or duplicates.

Use recursive CTEs when you need to explore hierarchical or connected data stepwise; for simple joins, recursion is not needed.

Summary

Recursive CTEs let you explore connected data step-by-step.

They have an anchor part and a recursive part that references itself.

Useful for graphs, trees, and hierarchical data traversal.

Practice

1. What is the main purpose of using a WITH RECURSIVE clause in PostgreSQL when working with graph data?

easy

A. To optimize query performance by indexing graph nodes

B. To create temporary tables for storing graph data

C. To repeatedly execute a query that references itself to explore connected nodes

D. To join two unrelated tables without conditions

Solution

1. Step 1: Understand the purpose of recursive CTEs

   Recursive CTEs allow a query to call itself repeatedly, which is useful for exploring connected data like graph nodes.

2. Step 2: Identify the correct use case

   Using WITH RECURSIVE lets you traverse graph structures step-by-step by referencing the CTE inside itself.

3. Final Answer:

   To repeatedly execute a query that references itself to explore connected nodes -> Option C

4. Quick Check:

   Recursive CTE = self-referencing query [OK]

Hint: Recursive CTEs repeat queries to explore connected data [OK]

Common Mistakes:

• Confusing recursive CTE with temporary tables
• Thinking it automatically optimizes performance
• Using it for unrelated table joins

2. Which of the following is the correct basic syntax to start a recursive CTE named path in PostgreSQL?

easy

A. WITH path RECURSIVE AS (SELECT ... UNION ALL SELECT ... FROM path ...)

B. CREATE RECURSIVE TABLE path AS SELECT ...

C. SELECT RECURSIVE * FROM path

D. WITH RECURSIVE path AS (SELECT ... UNION ALL SELECT ... FROM path ...)

Solution

1. Step 1: Recall the syntax for recursive CTEs

   The correct syntax starts with WITH RECURSIVE cte_name AS ( ... ), where the CTE references itself inside.

2. Step 2: Match the syntax options

   WITH RECURSIVE path AS (SELECT ... UNION ALL SELECT ... FROM path ...) correctly uses WITH RECURSIVE path AS (SELECT ... UNION ALL SELECT ... FROM path ...). Other options misuse keywords or order.

3. Final Answer:

   WITH RECURSIVE path AS (SELECT ... UNION ALL SELECT ... FROM path ...) -> Option D

4. Quick Check:

   WITH RECURSIVE + CTE name + self-reference [OK]

Hint: Start with WITH RECURSIVE cte_name AS (...) [OK]

Common Mistakes:

• Swapping RECURSIVE and CTE name order
• Using CREATE instead of WITH
• Omitting UNION ALL between parts

3. Given the table edges(source INT, target INT) with data:
(1, 2), (2, 3), (3, 4), (2, 5)
What will the following query return?

WITH RECURSIVE path AS (SELECT source, target FROM edges WHERE source = 1 UNION ALL SELECT e.source, e.target FROM edges e JOIN path p ON e.source = p.target) SELECT * FROM path ORDER BY source, target;

medium

A. Rows: (1,2), (2,3), (2,5), (3,4)

B. Rows: (1,2), (2,3), (3,4)

C. Rows: (1,2), (2,5)

D. Empty result

Solution

1. Step 1: Identify the anchor and recursive parts

   The anchor selects edges where source=1: (1,2). The recursive part joins edges where source matches previous target.

2. Step 2: Trace recursive expansion

   From (1,2), next edges with source=2 are (2,3) and (2,5). From (2,3), next edge with source=3 is (3,4). No further edges from 5 or 4.

3. Final Answer:

   Rows: (1,2), (2,3), (2,5), (3,4) -> Option A

4. Quick Check:

   Recursive join expands all reachable edges [OK]

Hint: Follow edges from start node stepwise [OK]

Common Mistakes:

• Ignoring multiple edges from same source
• Stopping recursion too early
• Misordering results

4. Consider this recursive CTE:

WITH RECURSIVE path(node) AS (SELECT 1 UNION ALL SELECT node + 1 FROM path WHERE node < 3) SELECT * FROM path;

What is wrong if this query runs forever without stopping?

medium

A. Not selecting from the CTE in the final SELECT

B. Missing a proper termination condition in the recursive part

C. Using UNION instead of UNION ALL

D. Incorrect anchor query syntax

Solution

1. Step 1: Analyze the recursive condition

   The recursive part adds 1 to node while node < 3, but due to UNION ALL accumulating duplicates of rows satisfying the condition, recursion continues indefinitely.

2. Step 2: Identify why it loops infinitely

   If the termination condition is incorrect or not reached, recursion never stops. Here, the condition is present but ineffective against duplicates, causing infinite loop.

3. Final Answer:

   Missing a proper termination condition in the recursive part -> Option B

4. Quick Check:

   Infinite recursion = no stop condition [OK]

Hint: Check recursive WHERE clause stops recursion [OK]

Common Mistakes:

• Confusing UNION and UNION ALL effects
• Ignoring anchor query correctness
• Forgetting to select from CTE

5. You have a table employees(id INT, manager_id INT) representing a company hierarchy. How would you write a recursive CTE to find all employees under manager with id = 10, including indirect reports?

hard

A. WITH RECURSIVE subordinates AS (
SELECT id FROM employees WHERE manager_id = 10
UNION ALL
SELECT e.id FROM employees e JOIN subordinates s ON e.manager_id = s.id
) SELECT * FROM subordinates;

B. WITH RECURSIVE subordinates AS (
SELECT id FROM employees WHERE id = 10
UNION ALL
SELECT e.id FROM employees e JOIN subordinates s ON e.id = s.manager_id
) SELECT * FROM subordinates;

C. WITH RECURSIVE subordinates AS (
SELECT id FROM employees WHERE manager_id = 10
UNION ALL
SELECT e.manager_id FROM employees e JOIN subordinates s ON e.id = s.id
) SELECT * FROM subordinates;

D. WITH RECURSIVE subordinates AS (
SELECT id FROM employees WHERE manager_id = 10
UNION ALL
SELECT e.id FROM employees e JOIN subordinates s ON e.id = s.manager_id
) SELECT * FROM subordinates;

Solution

1. Step 1: Identify the anchor query

   The anchor selects employees directly reporting to manager 10: WHERE manager_id = 10.

2. Step 2: Understand the recursive join

   The recursive part joins employees whose manager_id matches any previously found employee id, thus finding indirect reports.

3. Final Answer:

   WITH RECURSIVE subordinates AS ( SELECT id FROM employees WHERE manager_id = 10 UNION ALL SELECT e.id FROM employees e JOIN subordinates s ON e.manager_id = s.id ) SELECT * FROM subordinates; -> Option A

4. Quick Check:

   Join on e.manager_id = s.id to find subordinates [OK]

Hint: Join on manager_id = previous employee id to find reports [OK]

Common Mistakes:

• Swapping join conditions (id vs manager_id)
• Starting anchor with wrong WHERE clause
• Selecting wrong columns in recursive part