我的组织结构如下:

      T1
|''''   ''''|
T2          T3
|
T4


在数据库中存储为-

+----+---------+-----------+-----------+
| ID | TEAM_ID | PARENT_ID | TEAM_NAME |
+----+---------+-----------+-----------+
|  1 |       1 |         1 | T1        |
|  2 |       2 |         2 | T2        |
|  3 |       2 |         1 | T2        |
|  4 |       3 |         1 | T3        |
|  5 |       3 |         3 | T3        |
|  6 |       4 |         4 | T4        |
|  7 |       4 |         2 | T4        |
|  8 |       4 |         1 | T4        |
+----+---------+-----------+-----------+


我想根据上表中提供的平面数据重新构建上述树。

我目前的做法是-

Map<Long, List<TeamHierarchy>> tree = new HashMap<>();
        for (TeamHierarchy n : flatTeamStructure) {
            if (n.getParentTeamId() == n.getTeamId()) {
                if (!tree.containsKey(n.getParentTeamId())) {
                    tree.put(n.getParentTeamId(), new ArrayList<TeamHierarchy>());
                }
            } else {
                if (!tree.containsKey(n.getParentTeamId())) {
                    tree.put(n.getParentTeamId(), new ArrayList<TeamHierarchy>());
                }
                tree.get(n.getParentTeamId()).add(n);
            }
        }


这并不完全正确,因为我在T1的孩子中也得到了T4。
我只想有直子。没有递归的任何建议都会有所帮助。

最佳答案

我不确定这是否是最有效的方法,但是它应该可以工作。我会尝试将每个团队ID映射到其适当的父对象。这里的困难是您的表包含多余的信息,因此您必须能够清除它。

这个想法是从根开始构建树,如果发现更深的树,则递归地修改父树。这是一个简单的独立程序示例,应该可以帮助您。

public class TestTree {
    private static List<Entry> entries = new ArrayList<Entry>();

    public static void main(String[] args) throws Exception {
        // simulate the DB entries
        entries.add(new Entry(1, 1, 1, "T1"));
        entries.add(new Entry(2, 2, 2, "T2"));
        entries.add(new Entry(3, 2, 1, "T2"));
        entries.add(new Entry(4, 3, 1, "T3"));
        entries.add(new Entry(5, 3, 3, "T3"));
        entries.add(new Entry(6, 4, 4, "T4"));
        entries.add(new Entry(7, 4, 2, "T4"));
        entries.add(new Entry(8, 4, 1, "T4"));

        // the root is the one entry with no parent other than self
        int root = 1;

        // map all relationships to the root
        Map<Integer, Integer> tree = new HashMap<Integer, Integer>();   // ID -> parent ID
        buildTree(tree, root);

        System.out.println(tree);

        // From this Map, it should be pretty obvious how to build the tree.
    }

    private static void buildTree(Map<Integer, Integer> tree, int parentId) {
        boolean dirty = false;
        for(Entry entry : entries) {
            if(entry.parentId == parentId && entry.teamId != parentId) {
                tree.put(entry.teamId, parentId);
                dirty = true;
            }
        }

        if(dirty) {
            // Continue building the tree from each node that was updated
            for(Integer nodeId : tree.keySet()) {
                if(tree.get(nodeId) == parentId) buildTree(tree, nodeId);
            }
        }
    }

    private static class Entry {
        int id;
        int teamId;
        int parentId;
        String teamName;

        Entry(int id, int teamId, int parentId, String teamName) {
            this.id = id;
            this.teamId = teamId;
            this.parentId = parentId;
            this.teamName = teamName;
        }
    }


更新资料

对于讨厌的递归方法(如果您的树太深了,以至于炸毁了方法调用堆栈):

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Stack;

public class TestTree {
    private static List<Entry> entries = new ArrayList<Entry>();

    public static void main(String[] args) throws Exception {
        // simulate the DB entries
        entries.add(new Entry(1, 1, 1, "T1"));
        entries.add(new Entry(2, 2, 2, "T2"));
        entries.add(new Entry(3, 2, 1, "T2"));
        entries.add(new Entry(4, 3, 1, "T3"));
        entries.add(new Entry(5, 3, 3, "T3"));
        entries.add(new Entry(6, 4, 4, "T4"));
        entries.add(new Entry(7, 4, 2, "T4"));
        entries.add(new Entry(8, 4, 1, "T4"));

        // the root is the one entry with no parent other than self
        int root = 1;

        // map all relationships to the root
        Map<Integer, Integer> tree = new HashMap<Integer, Integer>();   //    ID -> parent ID
        Stack<Integer> stack = new Stack<Integer>();

        stack.push(root);
        do {
            int parentId = stack.pop();

            if(buildTree(tree, parentId)) {
                // Continue building the tree from each node that was updated
                for(Integer nodeId : tree.keySet()) {
                    if(tree.get(nodeId) == parentId) stack.push(nodeId);
                }
            }
        } while(!stack.isEmpty());

        System.out.println(tree);

        // From this Map, it should be pretty obvious how to build the tree.
    }

    private static boolean buildTree(Map<Integer, Integer> tree, int parentId) {
        boolean dirty = false;
        for(Entry entry : entries) {
            if(entry.parentId == parentId && entry.teamId != parentId) {
                tree.put(entry.teamId, parentId);
                dirty = true;
            }
        }

        return dirty;
    }

    private static class Entry {
        int id;
        int teamId;
        int parentId;
        String teamName;

        Entry(int id, int teamId, int parentId, String teamName) {
            this.id = id;
            this.teamId = teamId;
            this.parentId = parentId;
            this.teamName = teamName;
        }
    }
}

关于java - Java中将组织结构平移到树状 View ,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/38528636/

10-10 19:31