给定一个节点表（它们的 Id 和它们的父 Id），我如何获取所有根节点、内部节点和叶节点？
这是我目前所拥有的：
根节点
SELECT Id, "Root" FROM NodeTable
WHERE ParentId IS NULL;
内部节点
???
叶节点
SELECT N1.Id, "Leaf" FROM NodeTable N1
LEFT JOIN NodeTable N2 ON N1.Id = N2.ParentId
WHERE N2.ParentId IS NULL;
这是正确的吗？有没有一种方法可以在一个查询中做到这一点？
【问题讨论】：
标签： mysql binary-tree
select case
when p is null then concat(n, ' Root')
when n in (select distinct p from bst) then concat(n, ' Inner')
else concat(n, ' Leaf')
end
from bst
order by n asc
【讨论】：
您的“内部节点”查询可以类似于您的“叶节点”查询得出。
SELECT DISTINCT N1.Id, "Inner"
FROM NodeTable N1
JOIN NodeTable N2
ON N2.ParentId = N1.ID
WHERE N1.ParentId IS NOT NULL;
这将为您提供所有不是根节点的节点 (N1)（因为它们具有非空 ParentId）并且具有子节点 (N2)。
是的，您可以使用UNION 关键字在一个查询中完成所有这些操作。
SELECT Id, "Root"
FROM NodeTable
WHERE ParentId IS NULL
UNION
SELECT DISTINCT N1.Id, "Inner"
FROM NodeTable N1
JOIN NodeTable N2
ON N2.ParentId = N1.ID
WHERE N1.ParentId IS NOT NULL
UNION
SELECT N1.Id, "Leaf"
FROM NodeTable N1
LEFT JOIN NodeTable N2
ON N1.Id = N2.ParentId
WHERE N2.ParentId IS NULL;
【讨论】：
JOIN 为每个内部节点的每个子节点生成一行。
更简洁的方式：
SELECT
N,
IF(
P IS NULL,
'Root',
IF(
(
SELECT COUNT(*)
FROM BST
WHERE P=B.N
) > 0,
'Inner',
'Leaf'
)
)
FROM BST AS B
ORDER BY N;
或
SELECT
N,
IF(
P IS NULL,
'Root',
IF(
B.N IN (SELECT P FROM BST),
'Inner',
'Leaf'
)
)
FROM BST AS B
ORDER BY N;
【讨论】：
select n, (case when p is null then "Root" when n in (select p from bst) then "Inner" else "Leaf" end) from BST order by n;
【讨论】：
您的根叶逻辑似乎是正确的。内部通常是其他一切
如果您想在 1 个 SQL 代码中执行此操作，那么您可以试试这个
SELECT DISTINCT Id,
CASE WHEN Id NOT IN (SELECT DISTINCT ParentId FROM NodeTable WHERE ParentId IS NOT NULL )
THEN "Leaf"
WHEN ParentId IS NULL THEN "Root"
ELSE "Inner"
END AS NodeType
FROM NodeTable
ORDER BY Id
【讨论】：
选择 n，大小写 当 p 为空时，'Root' 当 n in (select p from bst) 然后 'Inner' 否则'叶子' 结尾 从 bst 按 n 排序；
【讨论】：
短而高效
SELECT N , CASE WHEN P is null THEN 'Root' ELSE
(IF (N IN (SELECT P FROM BST ), 'Inner','Leaf' ) ) END FROM BST ORDER BY N
【讨论】：
SELECT DISTINCT(N), 'Root' FROM BST WHERE P IS NULL UNION
SELECT DISTINCT(N), 'Leaf' FROM BST WHERE N NOT IN (SELECT
DISTINCT(P) FROM BST WHERE P IS NOT NULL) UNION
SELECT DISTINCT(N), 'Inner' FROM BST WHERE N IN (SELECT
DISTINCT(P) FROM BST) AND P IS NOT NULL;
WHERE
Root= Parent is null,
leaf= when a Particular node is not a parent for any node,
Inner= when a Particular node is a parent for any node
【讨论】：
这是另一种选择：
SELECT distinct N1.N,
Case
When N1.P is null then 'Root'
When N2.P is null then 'Leaf'
Else 'Inner'
End
FROM BST N1
LEFT JOIN BST N2
ON N1.N = N2.P
ORDER BY N1.n
【讨论】：
select t.N,
case
when t.P is null then 'Root'
when t.N in (select t1.P from BST t1 where t1.P = t.N) then 'Inner'
else 'Leaf'
end
from BST t
order by t.N
【讨论】：
select n,case when p in (select n from bst) then case when n in (select p from bst) then "Inner" else "Leaf" end else "Null" end from bst order by n ;
【讨论】：
我相信它来自这个链接：https://www.hackerrank.com/challenges/binary-search-tree-1/problem
这是你可以做的！
SELECT N, IF(P IS NULL, 'Root', IF((SELECT COUNT(*) FROM BST WHERE P=B.N)>0,'Inner','Leaf'))
FROM BST AS B
ORDER BY N;
【讨论】：
以下是我的解决方案：
select N,case when level=1 then 'Root'
when connect_by_isleaf=1 then 'Leaf'
when level>1 and connect_by_isleaf=0 then 'Inner'
end
from BST
start with p is null
connect by p=prior N
order by N;
【讨论】：
Select distinct(N),x
from(
select N,P,
case when P is null and N is not null then 'Root'
when N in (Select distinct(P)from BST where P is not null) then 'Inner'
else 'Leaf' end as x
from BST)a
order by N;
【讨论】：
select N, 'Root' from N1 where P is null
UNION
select N, 'Leaf' from N1 where N not in (select P from N1 where p is not
null)
UNION
select N , 'Inner' from N1 where N in
(select P from N1 where p NOT IN (select N from N1 where P is null))
order by n;
【讨论】：
你也可以试试..
选择 N， B.p 为 NULL 然后为 'Root' 的情况 当 B.N IN（从 BST 中选择 p）然后 'Inner' 否则'叶子' 结束案例 从 BST B 按 N 排序；
【讨论】：
你可以使用存在和不存在：
选择 N，“根”
来自 BST，其中 p 为空
联盟
选择 N,'叶子'
来自 BST
不存在的地方（从 BST b 中选择 * b.P =a.N ）
工会
选择 N,'内部'
来自 BST
where exists (select * from BST b where b.P =a.N and a.P is not null);
【讨论】：
选择 n ，当 p 为空时，则为 'Root' 当 n in (select distinct p from bst ) 然后 'Inner' else 'Leaf' 以 t 结尾
从 bst 按 1 排序；
【讨论】：
SELECT N,CASE WHEN ISNULL(P,0) = 0 THEN 'Root'
WHEN N IN (SELECT P FROM BST ) THEN 'Inner'
ELSE 'Leaf'
END NODETYPE
FROM BST
ORDER BY N
【讨论】：
select N, 'Root' from BST where P is null
UNION
select N, 'Leaf' from BST where N not in (select P from BST where p is not
null)
UNION
select N , 'Inner' from BST where N in
(select P from BST where p NOT IN (select N from BST where P is null))
order by n;
【讨论】：
其他方式：
SELECT
N,
DECODE(P, NULL, 'Root', DECODE(N, (SELECT DISTINCT S.P FROM BST S WHERE S.P=B.N),'Inner', 'Leaf')) FROM BST B
ORDER BY N;
【讨论】：
select N, Case
when P is null then 'Root'
when N in (select P from BST) then 'Inner'
Else 'Leaf' end from BST order by N asc
【讨论】：