java树形结构 算法
最近看到一个有意思的树形结构，为每个节点添加了lft和rgt两个属性。这样查找该节点的子节点、查找该节点所有父节点，就不用去递归查询，只需要用between、and语句就可以实现。下面以创建一个栏目树为例，以下是我的理解。
一般来讲，我们创建栏目树的时候，大多只需要一个外键parentid来区分该节点属于哪个父节点。数据库的设计如下图：
这样一来，
1.查找该节点的所有子节点，则需要采用sql的递归语句：
select * from tableName connect by prior id=sj_parent_id start with id=1
(oracle 写法，mysql目前不支持，如果mysql想查找树形，可以利用存储过程).
2.查找该节点的父节点的sql递归语句：
select * from tableName connect by prior sj_parent_id =id start with id=1
如果数据量过大或者层次太多，那么这样操作是会影响性能的。
“任何树形结构都可以用二叉树来表示”。其实我们创建的栏目树就是一个简型的二叉树。根据数据结构里面二叉树的遍历，再稍微修改下，将数据库设计如下图所示：
这样我们查找该节点的所有子节点，则只需要查找id在lft和rgt之间的所有节点即可。
1.查找该节点的所有子节点的Sql语句为：
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select * from tb_subject s,tb_subject t where s.lft between t.lft and t.rgt and t.id=1
2.查找该节点的所有父节点的sql语句为：
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select s.* from tb_subject s,tb_subject t where s.lft<t.lft and (s.rgt-s.lft)>1 and s.rgt>t.rgt and t.id=1
下面来详细讲解下，怎么用java来实现这种算法。
<!--EndFragment-->1. 新增节点
新增节点比较简单，基本步骤为
A． 查找当前插入节点的父节点的lft值
B． 将树形中所有lft和rgt节点大于父节点左值的节点都+2
C． 将父节点左值+1，左值+2分别作为当前节点的lft和rgt
因为项目中采用的是struts2+hibernate3.2+spring2.5的框架，代码如下：
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public boolean onSave(Object entity, Serializable id, Object[] state,
String[] propertyNames, Type[] types) {
if (entity instanceof HibernateTree) {
HibernateTree tree = (HibernateTree) entity;
Long parentId = tree.getParentId();
String beanName = tree.getClass().getName();
Session session = getSession();
FlushMode model = session.getFlushMode();
session.setFlushMode(FlushMode.MANUAL);
Integer myPosition = new Integer(0);
//查找父节点的左值
if (parentId != null) {
String hql = "select b.lft from " + beanName
+ " b where b.id=:pid";
myPosition = (Integer) session.createQuery(hql).setLong("pid",
parentId).uniqueResult();
}
//将树形结构中所有大于父节点左值的右节点+2
String hql1 = "update " + beanName
+ " b set b.rgt = b.rgt + 2 WHERE b.rgt > :myPosition";
//将树形结构中所有大于父节点左值的左节点+2
String hql2 = "update " + beanName
+ " b set b.lft = b.lft + 2 WHERE b.lft > :myPosition";
if (!StringUtils.isBlank(tree.getTreeCondition())) {
hql1 += " and (" + tree.getTreeCondition() + ")";
hql2 += " and (" + tree.getTreeCondition() + ")";
}
session.createQuery(hql1).setInteger("myPosition", myPosition)
.executeUpdate();
session.createQuery(hql2).setInteger("myPosition", myPosition)
.executeUpdate();
session.setFlushMode(model);
//定位自己的左值(父节点左值+1)和右值(父节点左值+2)
for (int i = 0; i < propertyNames.length; i++) {
if (propertyNames[i].equals(HibernateTree.LFT)) {
state[i] = myPosition + 1;
}
if (propertyNames[i].equals(HibernateTree.RGT)) {
state[i] = myPosition + 2;
}
}
return true;
}
return false;
}
２. 修改节点
修改的时候比较麻烦，具体步骤为：
在修改lft和rgt之前，当前节点的父节点id已经改变
a. 查出当前节点的左右节点（nodelft、nodergt），并nodergt-nodelft+1 = span，获取父节点的左节点parentlft
b. 将所有大于parentlft的lft(左节点)、rgt(右节点)的值+span
c. 查找当前节点的左右节点（nodelft、nodergt），并parentlft-nodelft+1 = offset
d. 将所有lft(左节点) between nodelft and nodergt的值+offset
e. 将所有大于nodergt的lft(左节点)、rgt(右节点)的值-span
Java代码如下：
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public void updateParent(HibernateTree tree, HibernateTree preParent,
HibernateTree curParent) {
if (preParent != null && preParent != null
&& !preParent.equals(curParent)) {
String beanName = tree.getClass().getName();
// 获得节点位置
String hql = "select b.lft,b.rgt from " + beanName
+ " b where b.id=:id";
Object[] position = (Object[]) super.createQuery(hql).setLong(
"id", tree.getId()).uniqueResult();
System.out.println(hql+"| id = "+tree.getId());
int nodeLft = ((Number) position[0]).intValue();
int nodeRgt = ((Number) position[1]).intValue();
int span = nodeRgt - nodeLft + 1;
// 获得当前父节点左位置
hql = "select b.lft from " + beanName + " b where b.id=:id";
int parentLft = ((Number) super.createQuery(hql).setLong("id",
curParent.getId()).uniqueResult()).intValue();
System.out.println(hql+"| id = "+curParent.getId());
// 先空出位置
String hql1 = "update " + beanName + " b set b.rgt = b.rgt + "
+ span + " WHERE b.rgt > :parentLft";
String hql2 = "update " + beanName + " b set b.lft = b.lft + "
+ span + " WHERE b.lft > :parentLft";
if (!StringUtils.isBlank(tree.getTreeCondition())) {
hql1 += " and (" + tree.getTreeCondition() + ")";
hql2 += " and (" + tree.getTreeCondition() + ")";
}
super.createQuery(hql1).setInteger("parentLft", parentLft)
.executeUpdate();
super.createQuery(hql2).setInteger("parentLft", parentLft)
.executeUpdate();
System.out.println(hql1+"| parentLft = "+parentLft);
System.out.println(hql2+"| parentLft = "+parentLft);
// 再调整自己
hql = "select b.lft,b.rgt from " + beanName + " b where b.id=:id";
position = (Object[]) super.createQuery(hql).setLong("id",
tree.getId()).uniqueResult();
System.out.println(hql+"| id = "+tree.getId());
nodeLft = ((Number) position[0]).intValue();
nodeRgt = ((Number) position[1]).intValue();
int offset = parentLft - nodeLft + 1;
hql = "update "
+ beanName
+ " b set b.lft=b.lft+:offset, b.rgt=b.rgt+:offset WHERE b.lft between :nodeLft and :nodeRgt";
if (!StringUtils.isBlank(tree.getTreeCondition())) {
hql += " and (" + tree.getTreeCondition() + ")";
}
super.createQuery(hql).setParameter("offset", offset)
.setParameter("nodeLft", nodeLft).setParameter("nodeRgt",
nodeRgt).executeUpdate();
System.out.println(hql+"| offset = "+offset+" | nodelft = "+nodeLft+" | nodergt = "+ nodeRgt);
// 最后删除（清空位置）
hql1 = "update " + beanName + " b set b.rgt = b.rgt - " + span
+ " WHERE b.rgt > :nodeRgt";
hql2 = "update " + beanName + " b set b.lft = b.lft - " + span
+ " WHERE b.lft > :nodeRgt";
if (tree.getTreeCondition() != null) {
hql1 += " and (" + tree.getTreeCondition() + ")";
hql2 += " and (" + tree.getTreeCondition() + ")";
}
super.createQuery(hql1).setParameter("nodeRgt", nodeRgt)
.executeUpdate();
super.createQuery(hql2).setParameter("nodeRgt", nodeRgt)
.executeUpdate();
System.out.println(hql1+"| nodeRgt = "+nodeRgt);
System.out.println(hql2+"| nodeRgt = "+nodeRgt);
}
}
３. 删除节点
删除节点也比较简单，具体步骤为：
A． 查找要删除节点的lft值
B． 将所有lft和rgt大于删除节点lft值的都-2
Java代码如下：
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public void onDelete(Object entity, Serializable id, Object[] state,
String[] propertyNames, Type[] types) {
if (entity instanceof HibernateTree) {
HibernateTree tree = (HibernateTree) entity;
String beanName = tree.getClass().getName();
Session session = getSession();
FlushMode model = session.getFlushMode();
session.setFlushMode(FlushMode.MANUAL);
//查找要删除的节点的左值
String hql = "select b.lft from " + beanName + " b where b.id=:id";
Integer myPosition = (Integer) session.createQuery(hql).setLong(
"id", tree.getId()).uniqueResult();
//将所有大于删除节点左值的rgt都-2
String hql1 = "update " + beanName
+ " b set b.rgt = b.rgt - 2 WHERE b.rgt > :myPosition";
//将所有大于删除节点左值的lft都-2
String hql2 = "update " + beanName
+ " b set b.lft = b.lft - 2 WHERE b.lft > :myPosition";
if (tree.getTreeCondition() != null) {
hql1 += " and (" + tree.getTreeCondition() + ")";
hql2 += " and (" + tree.getTreeCondition() + ")";
}
session.createQuery(hql1).setInteger("myPosition", myPosition)
.executeUpdate();
session.createQuery(hql2).setInteger("myPosition", myPosition)
.executeUpdate();
session.setFlushMode(model);
}
}
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