MYSQL 8-如何确定“节点”是否为'；叶节点'；在CTE中检索树时

我有以下数据：

mysql> SELECT id, parent_id, name FROM account_type WHERE account_classification_id = 1 LIMIT 7;
+----+-----------+----------------------+
| id | parent_id | name                 |
+----+-----------+----------------------+
| 30 |      NULL | AKTIVA               |
| 40 |        30 | Aktiva Lancar        |
| 41 |        40 | Kas & Bank           |
| 42 |        41 | Kas                  |
| 43 |        41 | Kas Tunai USD        |
| 44 |        41 | Kas Tunai Di Brankas |
| 46 |        41 | Bank                 |
+----+-----------+----------------------+
7 rows in set (0.00 sec)

因此，对于CTE：

WITH RECURSIVE account_path (root, id, parent_id, name, lvl, `sort`, account_roll_up_id) AS
                   (
                       SELECT
                              id AS root,
                              id,
                              parent_id,
                              `name`, 0 lvl,
                              `sort`,
                              account_roll_up_id

                       FROM account_type
                                WHERE id IN (30, 213)
                       UNION ALL

                       SELECT
                              ap.root,
                              at.id,
                              at.parent_id,
                              at.name,
                              (ap.lvl + 1),
                              at.sort,
                              at.account_roll_up_id

                       FROM account_path AS ap
                                JOIN account_type AS at ON ap.id = at.parent_id
                   )

SELECT * FROM `account_path`
ORDER BY account_path.sort
;

我需要每一行都有一个布尔值，当我检索所有节点时，它被确定为一个叶子节点， 可能吗？ 大概是这样的：

+------------+------+------+-----------+----------------------+------+------+--------------------+
|  is_a_leaf | root | id   | parent_id  | name                 | lvl  | sort| account_roll_up_id |
+------------+------+------+------------+----------------------+------+-----+--------------------+
|       FALSE|   30 |   30 |      NULL | AKTIVA               |    0 |    1 |               NULL |
|       FALSE|   30 |   40 |        30 | Aktiva Lancar        |    1 |    2 |                  2 |
|       FALSE|   30 |   41 |        40 | Kas & Bank           |    2 |    3 |                  2 |
|       TRUE |   30 |   42 |        41 | Kas                  |    3 |    4 |                  2 |
|       TRUE |   30 |   43 |        41 | Kas Tunai USD        |    3 |    5 |                  2 |
|       TRUE |   30 |   44 |        41 | Kas Tunai Di Brankas |    3 |    6 |                  2 |
|       TRUE |   30 |   46 |        41 | Bank                 |    3 |    7 |                  2 |
+------------+------+------+-----------+----------------------+------+------+--------------------+

解决了，谢谢@Akina

WITH RECURSIVE account_path (is_a_leaf, root, id, parent_id, name, lvl, `sort`, account_roll_up_id) AS
                   (
                       SELECT id    is_a_leaf,
                              id AS root,
                              id,
                              parent_id,
                              `name`,
                              0     lvl,
                              `sort`,
                              account_roll_up_id

                       FROM account_type
                       WHERE id IN (30, 213)
                       UNION ALL

                       SELECT (SELECT EXISTS(
                                              SELECT parent_id
                                              FROM account_type
                                              WHERE parent_id = at.id
                                          )),
                              ap.root,
                              at.id,
                              at.parent_id,
                              at.name,
                              (ap.lvl + 1),
                              at.sort,
                              at.account_roll_up_id

                       FROM account_path AS ap
                                JOIN account_type AS at ON ap.id = at.parent_id
                   )

SELECT *
FROM `account_path`
ORDER BY account_path.sort
LIMIT 7;

“叶痕”是指没有孩子<代码>选择。。不存在（{检索childs的子查询}）作为IsLeaf..

WITH RECURSIVE account_path (is_a_leaf, root, id, parent_id, name, lvl, `sort`, account_roll_up_id) AS
                   (
                       SELECT id    is_a_leaf,
                              id AS root,
                              id,
                              parent_id,
                              `name`,
                              0     lvl,
                              `sort`,
                              account_roll_up_id

                       FROM account_type
                       WHERE id IN (30, 213)
                       UNION ALL

                       SELECT (SELECT EXISTS(
                                              SELECT parent_id
                                              FROM account_type
                                              WHERE parent_id = at.id
                                          )),
                              ap.root,
                              at.id,
                              at.parent_id,
                              at.name,
                              (ap.lvl + 1),
                              at.sort,
                              at.account_roll_up_id

                       FROM account_path AS ap
                                JOIN account_type AS at ON ap.id = at.parent_id
                   )

SELECT *
FROM `account_path`
ORDER BY account_path.sort
LIMIT 7;