我有一个部门关系表如下：
+---------------+----------------+
| Dept. | superior Dept. |
+---------------+----------------+
| "00-01" | "00" |
| "00-02" | "00" |
| "00-01-01" | "00-01" |
| "00-01-02" | "00-01" |
| "00-01-03" | "00-01" |
| "00-02-01" | "00-02" |
| "00-02-03" | "00-02" |
| "00-02-03-01" | "00-02-03" |
+---------------+----------------+
现在我想根据他们的等级列出它们：
+-----------+--------------+--------------+--------------+
| Top Dept. | 2-tier Dept. | 3-tire Dept. | 4-tier Dept. |
+-----------+--------------+--------------+--------------+
| 00 | | | |
| | 00-01 | | |
| | | 00-01-01 | |
| | | 00-01-02 | |
| | 00-02 | | |
| | | 00-02-01 | |
| | | 00-02-03 | |
| | | | 00-02-03-01 |
+-----------+--------------+--------------+--------------+
我可以使用下面的代码构建关系树：
TreeNode.java
import java.util.LinkedList;
import java.util.List;
public class TreeNode {
public String value;
public List children = new LinkedList();
public TreeNode(String rootValue) {
value = rootValue;
}
}
PairsToTree.java
import java.util.*;
public class PairsToTree {
public static void main(String[] args) throws Exception {
// Create the child to parent hash map
Map <String, String> childParentMap = new HashMap<String, String>(8);
childParentMap.put("00-01", "00");
childParentMap.put("00-02", "00");
childParentMap.put("00-01-01", "00-01");
childParentMap.put("00-01-02", "00-01");
childParentMap.put("00-01-03", "00-01");
childParentMap.put("00-02-01", "00-02");
childParentMap.put("00-02-03", "00-02");
childParentMap.put("00-02-03-01", "00-02-03");
// All children in the tree
Collection<String> children = childParentMap.keySet();
// All parents in the tree
Collection<String> values = childParentMap.values();
// Using extra space here as any changes made to values will
// directly affect the map
Collection<String> clonedValues = new HashSet();
for (String value : values) {
clonedValues.add(value);
}
// Find parent which is not a child to any node. It is the
// root node
clonedValues.removeAll(children);
// Some error handling
if (clonedValues.size() != 1) {
throw new Exception("More than one root found or no roots found");
}
String rootValue = clonedValues.iterator().next();
TreeNode root = new TreeNode(rootValue);
HashMap<String, TreeNode> valueNodeMap = new HashMap();
// Put the root node into value map as it will not be present
// in the list of children
valueNodeMap.put(root.value, root);
// Populate all children into valueNode map
for (String child : children) {
TreeNode valueNode = new TreeNode(child);
valueNodeMap.put(child, valueNode);
}
// Now the map contains all nodes in the tree. Iterate through
// all the children and
// associate them with their parents
for (String child : children) {
TreeNode childNode = valueNodeMap.get(child);
String parent = childParentMap.get(child);
TreeNode parentNode = valueNodeMap.get(parent);
parentNode.children.add(childNode);
}
// Traverse tree in level order to see the output. Pretty
// printing the tree would be very
// long to fit here.
Queue q1 = new ArrayDeque();
Queue q2 = new ArrayDeque();
q1.add(root);
Queue<TreeNode> toEmpty = null;
Queue toFill = null;
while (true) {
if (false == q1.isEmpty()) {
toEmpty = q1;
toFill = q2;
} else if (false == q2.isEmpty()) {
toEmpty = q2;
toFill = q1;
} else {
break;
}
while (false == toEmpty.isEmpty()) {
TreeNode node = toEmpty.poll();
System.out.print(node.value + ", ");
toFill.addAll(node.children);
}
System.out.println("");
}
}
}
但无法理解 如何格式化输出以类似于表。或者是否有一个sql语句/存储过程（如this question）来执行此操作？
编辑：部门名称只是为了方便起见，它可以是任意字符串。
答案 0 :(得分：2)
您还没有注意到您使用的是哪种RDBMS，但是如果它恰好是MS SQL Server或某些版本的Oracle（请参阅下面的编辑），并且只是作为我们这些使用T-SQL并且感兴趣的人的服务在这个问题中，您可以使用recursive CTE在SQL Server中完成此任务。
编辑：Oracle（我的经验非常非常少）似乎也支持递归，包括自11g第2版以来的递归CTE。See this question for more information。鉴于此设置：
CREATE TABLE hierarchy
([Dept] varchar(13), [superiorDept] varchar(12))
;
INSERT INTO hierarchy
([Dept], [superiorDept])
VALUES
('00',NULL),
('00-01', '00'),
('00-02', '00'),
('00-01-01', '00-01'),
('00-01-02', '00-01'),
('00-01-03', '00-01'),
('00-02-01', '00-02'),
('00-02-03', '00-02'),
('00-02-03-01', '00-02-03')
;
这个递归CTE：
WITH cte AS (
SELECT
Dept,
superiorDept,
0 AS depth,
CAST(Dept AS NVARCHAR(MAX)) AS sort
FROM hierarchy h
WHERE superiorDept IS NULL
UNION ALL
SELECT
h2.Dept,
h2.superiorDept,
cte.depth + 1 AS depth,
CAST(cte.sort + h2.Dept AS NVARCHAR(MAX)) AS sort
FROM hierarchy h2
INNER JOIN cte ON h2.superiorDept = cte.Dept
)
SELECT
CASE WHEN depth = 0 THEN Dept END AS TopTier,
CASE WHEN depth = 1 THEN Dept END AS Tier2,
CASE WHEN depth = 2 THEN Dept END AS Tier3,
CASE WHEN depth = 3 THEN Dept END AS Tier4
FROM cte
ORDER BY sort
将呈现所需的输出：
TopTier Tier2 Tier3 Tier4
00
00-01
00-01-01
00-01-02
00-01-03
00-02
00-02-01
00-02-03
00-02-03-01
Here is a SQLFiddle to demonstrate
无论邻接列表中涉及的两列的内容或类型如何，这都应该有效。只要父值与id值匹配，CTE应该没有问题生成深度和排序信息。
完全在SQL中执行此操作的唯一警告是您必须手动定义一组数量的列，但在处理SQL时几乎总是如此。您可能希望生成CTE，然后将已排序的结果传递给您的程序并使用深度信息使列操作变得微不足道。
答案 1 :(得分：1)
这是一个基于您的数据的测试用例，以表格方式打印部门，尝试运行它，看看会发生什么
package test;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
public class DeptHierachy {
public Map<String,List<String>> deptMap= new HashMap<>();
public static void main(String[] args) {
TreeNode tn=new TreeNode("00");
tn.addChildTo("00-01", "00");
tn.addChildTo("00-02", "00");
tn.addChildTo("00-01-01", "00-01");
tn.addChildTo("00-01-02", "00-01");
tn.addChildTo("00-01-03", "00-01");
tn.addChildTo("00-02-01", "00-02");
tn.addChildTo("00-02-03", "00-02");
tn.addChildTo("00-02-03-01", "00-02-03");
tn.print();
//System.out.println("max height=" + tn.maxHeigth());
}
public static class TreeNode {
public String value;
public List<TreeNode> children = new LinkedList<>();
public TreeNode(String rootValue) {
value = rootValue;
}
public boolean addChildTo(String childName, String parentName) {
if (parentName.equals(value)) {
for (TreeNode child:children) {
if (child.getValue().equals(childName)) {
throw new IllegalArgumentException(parentName + " already has a child named " + childName);
}
}
children.add(new TreeNode(childName));
return true;
} else {
for (TreeNode child:children) {
if (child.addChildTo(childName, parentName)) {
return true;
}
}
}
return false;
}
public String getValue() {
return value;
}
public void print() {
int maxHeight=maxHeigth();
System.out.printf("|%-20.20s",value);
for (int i=0;i<maxHeight-1;i++) {
System.out.printf("|%-20.20s","");
}
System.out.println("|");
for (TreeNode child:children) {
child.print(1,maxHeight);
}
}
public void print(int level, int maxHeight) {
for (int i=0;i<level;i++) {
System.out.printf("|%-20.20s","");
}
System.out.printf("|%-20.20s",value);
for (int i=level;i<maxHeight-1;i++) {
System.out.printf("|%-20.20s","");
}
System.out.println("|");
for (TreeNode child:children) {
child.print(level+1,maxHeight);
}
}
public int maxHeigth() {
int localMaxHeight=0;
for (TreeNode child:children) {
int childHeight = child.maxHeigth();
if (localMaxHeight < childHeight) {
localMaxHeight = childHeight;
}
}
return localMaxHeight+1;
}
}
}
答案 2 :(得分：0)
浏览计算-个字符的值。最大-个字数+1是表列数。值大小是行数。
你可以定义一个矩阵，例如表格String[][]。
对于每个值，您可以找到列（在String值中计算-个字符）。
现在要查找行，您应该检查上一列以查找父String。
如果发现只是在找到的父行向下移动所有其余行之后插入一个新行。
如果未找到父String，则首先插入父项（基于childParentMap）。
如果parent为null，则只在末尾插入一个新行。
答案 3 :(得分：0)
import java.util.LinkedList;
import java.util.List;
public class TreeNode {
private String value;
private int depth;
private List children = new LinkedList<TreeNode>();
public TreeNode(String rootValue, int depth) {
value = rootValue;
this.depth = depth;
}
public boolean add(String parent, String child) {
if (this.value.equals(parent)) {
this.children.add(new TreeNode(child, depth + 1));
return true;
} else {
for (TreeNode node : children) {
if (node.add(parent, child)) {
return true;
}
}
}
return false;
}
public String print() {
StringBuilder sb = new StringBuilder();
sb.append(format());
for (TreeNode child : children) {
sb.append(child.print())
}
return sb.toString();
}
public String format() {
// Format this node according to depth;
}
}
首先打印Top Dept.，然后在Top Dept上调用print（）。