树结构数据查询父子级

数据库一般会提供树形结构的数据的查询函数，但是可能需要额外的配置，偶尔会因为设计上的失误，会导致写查询语句跟吃屎一样难受，数据量不大的情况，索性全查在代码中进行组合

import java.util.*;

/**
 * @author css
 * @data 2019/6/24 10:33
 */
public class Test<T> {
    private List<Node<T>> list = new ArrayList<>();

    public class Node<T> {
        Object id;
        Object pid;
        String name;
        T t;

        Node<T> parent;
        LinkedList<Node<T>> children = new LinkedList<>();

        @Override
        public String toString() {
            return "Node{" +
                    "id=" + id +
                    ", pid=" + pid +
                    ", name=" + name +
                    ", t=" + t +
                    ", children=" + children +
                    ‘}‘;
        }
    }

    public void add(Object id, Object pid, String name, T t) {
        Node<T> node = new Node<>();
        node.id = id;
        node.pid = pid;
        node.name = name;
        node.t = t;
        list.add(node);
    }

    public Node<T> toTree(Object root) {
        for (Node<T> i : list) {
            for (Node<T> j : list) {
                if (i.id.equals(j.pid)) {
                    i.children.add(j);
                    j.parent = i;
                }
            }
        }
        for (Node<T> i : list) {
            if (root.equals(i.id)) {
                return i;
            }
        }
        return null;
    }

    public List<T> getParent(Node<T> node) {
        Node<T> p = node.parent;
        if (p == null) {
            return null;
        } else {
            List<T> list = new ArrayList<>();
            while (p != null) {
                list.add(p.t);
                p = p.parent;
            }
            return list;
        }
    }

    public List<T> getChildren(Node<T> node) {
        LinkedList<Node<T>> c = new LinkedList<>(node.children);
        if (c.isEmpty()) {
            return null;
        } else {
            List<T> list = new ArrayList<>(c.size());
            Node<T> e;
            while (!c.isEmpty()) {
                e = c.pollFirst();
                list.add(e.t);
                if (!e.children.isEmpty()) {
                    c.addAll(e.children);
                }
            }
            return list;
        }
    }

    public static void main(String[] args) {
        Test<String> t = new Test<>();
        t.add(1, null, "1000", "node1");
        t.add(2, 1, "1001", "node2");
        t.add(3, 2, "1002", "node3");
        t.add(4, 2, "1003", "node4");
        Test.Node n = t.toTree(1);
        System.out.println(t.getChildren(n));
    }
}

原文地址：https://www.cnblogs.com/chenss15060100790/p/11122154.html