# Leetcode: Tree Node

Tree Node

Similar Problems:

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:

```	1
/   \
2       3
/   \
4       5
```

Note

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/tree-node
## Basic Idea: Left join
# In tree, each node can only one parent or no parent
## | id | p_id | id (child) |
## |----+------+------------|
## |  1 | null |          1 |
## |  1 | null |          2 |
## |  2 |    1 |          4 |
## |  2 |    1 |          5 |
## |  3 |    1 |       null |
## |  4 |    2 |       null |
## |  5 |    2 |       null |

select t1.id,
case
when isnull(t1.p_id) then 'Root'
when isnull(max(t2.id)) then 'Leaf'
else 'Inner'
end as Type
from tree as t1 left join tree as t2
on t1.id = t2.p_id
group by t1.id, t1.p_id
```

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