[SQL]602(根据属性判定Root/Inner/Leaf)+612(两两距离)

Posted 2021-03-01 wyz-2020

608.树节点

思路

三种分类点的特性：

Root：p_id为空；Inner：既有p_id又是别人的p_id；Leaf：剩下的。

所以代码：

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;

612. 平面上的最近距离

思路：

将两个表自联结，用POWER(,2)算距离，作为dist。甩掉dist=0的行(这里只能用HAVING)，而我暂时还不知原因。

排序取1.(LIMIT 1)

代码

SELECT ROUND(SQRT(POWER(p1.x - p2.x, 2)+POWER(p1.y - p2.y, 2)), 2) AS shortest
FROM point_2d p1, point_2d p2
HAVING shortest != 0
ORDER BY shortest
LIMIT 1;

如果非要用WHERE的话

SELECT ROUND(SQRT(POWER(p1.x - p2.x, 2)+POWER(p1.y - p2.y, 2)), 2) AS shortest
FROM point_2d p1, point_2d p2
WHERE (p1.x, p1.y) != (p2.x, p2.y)
ORDER BY shortest
LIMIT 1;