# Tree Node Problem

## Description

LeetCode Problem 608.

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

``````1
2
3
4
5
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7
8
9
+----+------+
| 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:

``````1
2
3
4
5
6
7
8
9
+----+------+
| 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:
``````1
2
3
4
5
1
/   \
2       3
/   \
4       5
``````

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

## MySQL Solution

``````1
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select id, 'Root' as Type
from tree
where p_id is null

union

select id, 'Leaf' as Type
from tree
where id not in (select distinct p_id
from tree
where p_id is not null)
and p_id is not null

union

select id, 'Inner' as Type
from tree
where id in (select distinct p_id
from tree
where p_id is not null)
and p_id is not null
order by id
``````