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Question
Solution

Given a table tree, id is identifier of the tree node and p_id is its parent node's id.

+----+------+
| id | p_id |
+----+------+
| 1  | null |
| 2  | 1    |
| 3  | 1    |
| 4  | 2    |
| 5  | 2    |
+----+------+

Each node in the tree can be one of three types:

Leaf: if the node is a leaf node.

Root: if the node is the root of the tree.

Inner: If the node is neither a leaf node nor a root node.

Write a query to print the node id and the type of the node. Sort your output by the node id. The result for the above sample is:

+----+------+
| id | Type |
+----+------+
| 1  | Root |
| 2  | Inner|
| 3  | Leaf |
| 4  | Leaf |
| 5  | Leaf |
+----+------+

Explanation

Node '1' is root node, because its parent node is NULL and it has child node '2' and '3'.

Node '2' is inner node, because it has parent node '1' and child node '4' and '5'.

Node '3', '4' and '5' is Leaf node, because they have parent node and they don't have child node.

And here is the image of the sample tree as below:

Note

If there is only one node on the tree, you only need to output its root attributes.

Solution
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Solution

Approach I: Using `UNION` [Accepted]

Intuition

We can print the node type by judging every record by its definition in this table.\n Root: it does not have a parent node at all\n Inner: it is the parent node of some nodes, and it has a not NULL parent itself.\n* Leaf: rest of the cases other than above two

Algorithm

By transiting the node type definition, we can have the following code.

For the root node, it does not have a parent.

SELECT\n    id, \'Root\' AS Type\nFROM\n    tree\nWHERE\n    p_id IS NULL\n

For the leaf nodes, they do not have any children, and it has a parent.

SELECT\n    id, \'Leaf\' AS Type\nFROM\n    tree\nWHERE\n    id NOT IN (SELECT DISTINCT\n            p_id\n        FROM\n            tree\n        WHERE\n            p_id IS NOT NULL)\n        AND p_id IS NOT NULL\n

For the inner nodes, they have have some children and a parent.

SELECT\n    id, \'Inner\' AS Type\nFROM\n    tree\nWHERE\n    id IN (SELECT DISTINCT\n            p_id\n        FROM\n            tree\n        WHERE\n            p_id IS NOT NULL)\n        AND p_id IS NOT NULL\n

So, one solution to the problem is to combine these cases together using UNION.

MySQL

SELECT\n    id, \'Root\' AS Type\nFROM\n    tree\nWHERE\n    p_id IS NULL\n\nUNION\n\nSELECT\n    id, \'Leaf\' AS Type\nFROM\n    tree\nWHERE\n    id NOT IN (SELECT DISTINCT\n            p_id\n        FROM\n            tree\n        WHERE\n            p_id IS NOT NULL)\n        AND p_id IS NOT NULL\n\nUNION\n\nSELECT\n    id, \'Inner\' AS Type\nFROM\n    tree\nWHERE\n    id IN (SELECT DISTINCT\n            p_id\n        FROM\n            tree\n        WHERE\n            p_id IS NOT NULL)\n        AND p_id IS NOT NULL\nORDER BY id;\n

Approach II: Using flow control statement `CASE` [Accepted]

Algorithm

The idea is similar with the above solution but the code is simpler by utilizing the flow control statements, which is effective to output differently based on different input values. In this case, we can use CASE statement.

MySQL

SELECT\n    id AS `Id`,\n    CASE\n        WHEN tree.id = (SELECT atree.id FROM tree atree WHERE atree.p_id IS NULL)\n          THEN \'Root\'\n        WHEN tree.id IN (SELECT atree.p_id FROM tree atree)\n          THEN \'Inner\'\n        ELSE \'Leaf\'\n    END AS Type\nFROM\n    tree\nORDER BY `Id`\n;\n

\n
MySQL provides different flow control statements besides CASE. You can try to rewrite the slution above using IF flow control statement.
\n

Approach III: Using `IF` function [Accepted]

Algorithm

Also, we can use a single IF function instead of the complex flow control statements.

MySQL

SELECT\n    atree.id,\n    IF(ISNULL(atree.p_id),\n        \'Root\',\n        IF(atree.id IN (SELECT p_id FROM tree), \'Inner\',\'Leaf\')) Type\nFROM\n \xc2\xa0 \xc2\xa0tree atree\nORDER BY atree.id\n

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Note: This solution was inspired by @richarddia
\n

608. Tree Node

Solution

Approach I: Using UNION [Accepted]

Approach II: Using flow control statement CASE [Accepted]

Approach III: Using IF function [Accepted]

Approach I: Using `UNION` [Accepted]

Approach II: Using flow control statement `CASE` [Accepted]

Approach III: Using `IF` function [Accepted]