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608. Tree Node

中文文档

Description

Table: Tree

+-------------+------+
| Column Name | Type |
+-------------+------+
| id          | int  |
| p_id        | int  |
+-------------+------+
id is the column with unique values for this table.
Each row of this table contains information about the id of a node and the id of its parent node in a tree.
The given structure is always a valid tree.

 

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 solution to report the type of each node in the tree.

Return the result table in any order.

The result format is in the following example.

 

Example 1:

Input: 
Tree table:
+----+------+
| id | p_id |
+----+------+
| 1  | null |
| 2  | 1    |
| 3  | 1    |
| 4  | 2    |
| 5  | 2    |
+----+------+
Output: 
+----+-------+
| id | type  |
+----+-------+
| 1  | Root  |
| 2  | Inner |
| 3  | Leaf  |
| 4  | Leaf  |
| 5  | Leaf  |
+----+-------+
Explanation: 
Node 1 is the root node because its parent node is null and it has child nodes 2 and 3.
Node 2 is an inner node because it has parent node 1 and child node 4 and 5.
Nodes 3, 4, and 5 are leaf nodes because they have parent nodes and they do not have child nodes.

Example 2:

Input: 
Tree table:
+----+------+
| id | p_id |
+----+------+
| 1  | null |
+----+------+
Output: 
+----+-------+
| id | type  |
+----+-------+
| 1  | Root  |
+----+-------+
Explanation: If there is only one node on the tree, you only need to output its root attributes.

Solutions

Solution 1: Conditional Statements + Subquery

We can use the CASE WHEN conditional statement to determine the type of each node as follows:

  • If a node's p_id is NULL, then it is a root node.
  • Otherwise, if a node is the parent node of another node (we use a subquery to determine this), then it is an internal node.
  • Otherwise, it is a leaf node.
# Write your MySQL query statement below
SELECT
    id,
    CASE
        WHEN p_id IS NULL THEN 'Root'
        WHEN id IN (SELECT p_id FROM Tree) THEN 'Inner'
        ELSE 'Leaf'
    END AS type
FROM Tree;