java 递归数据库生成 树形结构问题

1、准备表结构及对应的表数据

a、表结构：

create table TB_TREE

(

CID NUMBER not null,

CNAME VARCHAR2(50),

PID NUMBER //父节点

)

b、表数据：

insert into tb_tree (CID, CNAME, PID) values (1, '中国', 0)

insert into tb_tree (CID, CNAME, PID) values (2, '北京市', 1)

insert into tb_tree (CID, CNAME, PID) values (3, '广东省', 1)

insert into tb_tree (CID, CNAME, PID) values (4, '上海市', 1)

insert into tb_tree (CID, CNAME, PID) values (5, '广州市', 3)

insert into tb_tree (CID, CNAME, PID) values (6, '深圳市', 3)

insert into tb_tree (CID, CNAME, PID) values (7, '海珠区', 5)

insert into tb_tree (CID, CNAME, PID) values (8, '天河区', 5)

insert into tb_tree (CID, CNAME, PID) values (9, '福田区', 6)

insert into tb_tree (CID, CNAME, PID) values (10, '南山区', 6)

insert into tb_tree (CID, CNAME, PID) values (11, '密云县', 2)

insert into tb_tree (CID, CNAME, PID) values (12, '浦东', 4)

2、TreeNode对象，对应tb_tree

public class TreeNode implements Serializable {

private Integer cid

private String cname

private Integer pid

private List nodes = new ArrayList()

public TreeNode() {

}

//getter、setter省略

}

3、测试数据

public class TreeNodeTest {

@Test

public void loadTree() throws Exception{

System.out.println(JsonUtils.javaToJson(recursiveTree(1)))

}

/**

* 递归算法解析成树形结构

*

* @param cid

* @return

* @author jiqinlin

*/

public TreeNode recursiveTree(int cid) {

//根据cid获取节点对象(SELECT * FROM tb_tree t WHERE t.cid=?)

TreeNode node = personService.getreeNode(cid)

//查询cid下的所有子节点(SELECT * FROM tb_tree t WHERE t.pid=?)

List childTreeNodes = personService.queryTreeNode(cid)

//遍历子节点

for(TreeNode child : childTreeNodes){

TreeNode n = recursiveTree(child.getCid())//递归

node.getNodes().add(n)

}

return node

}

}

输出的json格式如下：

{

"cid": 1,

"nodes": [

{

"cid": 2,

"nodes": [

{

"cid": 11,

"nodes": [

],

"cname": "密云县",

"pid": 2

}

],

"cname": "北京市",

"pid": 1

},

{

"cid": 3,

"nodes": [

{

"cid": 5,

"nodes": [

{

"cid": 7,

"nodes": [

],

"cname": "海珠区",

"pid": 5

},

{

"cid": 8,

"nodes": [

],

"cname": "天河区",

"pid": 5

}

],

"cname": "广州市",

"pid": 3

},

{

"cid": 6,

"nodes": [

{

"cid": 9,

"nodes": [

],

"cname": "福田区",

"pid": 6

},

{

"cid": 10,

"nodes": [

],

"cname": "南山区",

"pid": 6

}

],

"cname": "深圳市",

"pid": 3

}

],

"cname": "广东省",

"pid": 1

},

{

"cid": 4,

"nodes": [

{

"cid": 12,

"nodes": [

],

"cname": "浦东",

"pid": 4

}

],

"cname": "上海市",

"pid": 1

}

],

"cname": "中国",

"pid": 0

}

二叉树

1

2

3

4

5

6

7

这个二叉树的深度是3，树的深度是最大结点所在的层，这里是3.

应该计算所有结点层数，选择最大的那个。

根据上面的二叉树代码，递归过程是：

f

(1)=f

(2)+1

>

f

(3)

+1

?

f(2)

+

1

:

f(3)

+1

f(2)

跟f(3)计算类似上面，要计算左右结点，然后取大者

所以计算顺序是f(4.left)

=

0,

f(4.right)

=

0

f

(4)

=

f(4.right)

+

1

=

1

然后计算f(5.left)

=

0,f(5.right)

=

0

f

(5)

=

f(5.right)

+

1

=1

f(2)

=

f(5)

+

1

=2

f(1.left)

计算完毕，计算f(1.right)

f(3)

跟计算f(2)的过程一样。

得到f(3)

=

f(7)

+1

=

2

f(1)

=

f(3)

+

1

=3

12345if(depleft>depright){return depleft+1}else{return depright+1}

只有left大于right的时候采取left

+1，相等是取right

如果数据库是oracle，可以用递归的sql实现

如果想用java实现

第一步遍历节点放入map结构

再次遍历节点，取出当前节点的父节点，parentNode.setchild(courrentNode)

这样第二次遍历完后已经是树形结构了。

从map中取出root节点就行

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