Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 ->
-- 查找所有父节点with tab as( select Type_Id,ParentId,Type_Name from Sys_ParamType_V2_0 where Type_Id=316--子节点 union all select b.Type_Id,b.ParentId,b.Type_Name from tab a,--子节点数据集 Sys_ParamType_V2_0 b --父节点数据集 where a.ParentId=b.Type_Id --子节点数据集.paren
如下一张表test:id name pid----------- ---------- -----------1 电器 NULL2 家电 13 冰箱 24 洗衣机 25 电脑 16 笔记本 57 平板 58 组装机 79 品牌机 7--查询电脑的所有子节点. 可采用标准sql的with实现递归查询: with subRecord(id,name,pid) as ( select id,name,pid from test where id = 5 union all select test.id
--------------------01.向上查找所有父节点-----------------WITH TEMP AS (SELECT * FROM CO_Department WHERE ID=11 --表的IDUNION ALL SELECT T0.* FROM TEMP,CO_Department T0 WHERE TEMP.ParentID=T0.ID --父级ID==子级ID)SELECT * FROM TEMP; 如图:根据"测试组"查找所有父节点 执行结果
2.3 Implement an algorithm to delete a node in the middle of a singly linked list, given only access to that node.EXAMPLEInput: the node c from the linked list a->b->c->d->eResult: nothing is returned, but the new linked list looks like a- >
mysql遍历节点的所有子节点 DELIMITER // CREATE FUNCTION `getChildrenList`(rootId INT) ) BEGIN ); ); SET sTemp = ''; SET sTempChd =CAST(rootId AS CHAR); WHILE sTempChd IS NOT NULL DO SET sTemp = CONCAT(sTemp,',',sTempChd); ; END WHILE; RETURN sTemp; END ); FIND_
T-Sql 递归查询(给定节点查所有父节点.所有子节点的方法) -- 查找所有父节点with tab as( select Type_Id,ParentId,Type_Name from Sys_ParamType_V2_0 where Type_Id=316--子节点 union all select b.Type_Id,b.ParentId,b.Type_Name from tab a,--子节点数据集 Sys_ParamType_V2_0 b --父节点数据集 where a.