Java list如何根据id获取子节点

工作中因业务需求，将数据库中的树状结构的数据根据父节点获取所有的子节点

实现思路

1.获取整个数据的list集合数据

2.将数据分组，java8 list有groupby分组，java8之前的自己遍历整理

3.分组后递归获取子节点，有子节点的添加，没有的设置子节点并删除分组的数据，知道分组数据删完

Tree.java

@Data
public class Tree {
 private Integer id;
 private Integer pId;
 private String key;
 private String value;
 private List<Tree> childList;
}

TreeUtils.java

public class TreeUtils {
 static List<Tree> trees ;
 static {
   String jsonStr = "[" +
       "{\"id\":100,\"pId\":1,\"key\":\"root\", \"value\": \"root\"}," +
       "{\"id\":1000,\"pId\":100,\"key\":\"node1\", \"value\": \"node1\"}," +
       "{\"id\":2000,\"pId\":100,\"key\":\"node2\",\"value\": \"node2\"}," +
       "{\"id\":3000,\"pId\":100,\"key\":\"node3\",\"value\": \"node3\"}," +
       "{\"id\":1100,\"pId\":1000,\"key\":\"node11\",\"value\": \"node11\"}," +
       "{\"id\":1200,\"pId\":1000,\"key\":\"node12\",\"value\": \"node12\"}," +
       "{\"id\":1110,\"pId\":1100,\"key\":\"node111\",\"value\": \"node111\"}," +
       "{\"id\":1120,\"pId\":1100,\"key\":\"node112\",\"value\": \"node112\"}," +
       "{\"id\":2100,\"pId\":2000,\"key\":\"node21\",\"value\": \"node21\"}," +
       "{\"id\":2200,\"pId\":2000,\"key\":\"node22\",\"value\": \"node22\"}," +
       "{\"id\":2110,\"pId\":2100,\"key\":\"node211\",\"value\": \"node21\"}" +
       "]";
   trees = JSONObject.parseArray(jsonStr, Tree.class);
 }

public static void main(String[] args) {
   Tree tree = metaTree(trees, 100);
   /**
    * Tree@6073f712[id=100,pId=1,key=root,value=root,childList=[
    *           Tree(id=1000, pId=100, key=node1, value=node1, childList=[
    *             Tree(id=1100, pId=1000, key=node11, value=node11, childList=[
    *               Tree(id=1110, pId=1100, key=node111, value=node111, childList=null),
    *               Tree(id=1120, pId=1100, key=node112, value=node112, childList=null)]),
    *             Tree(id=1200, pId=1000, key=node12, value=node12, childList=null)]),
    *           Tree(id=2000, pId=100, key=node2, value=node2, childList=[
    *             Tree(id=2100, pId=2000, key=node21, value=node21, childList=[
    *               Tree(id=2110, pId=2100, key=node211, value=node21, childList=null)]),
    *               Tree(id=2200, pId=2000, key=node22, value=node22, childList=null)]),
    *           Tree(id=3000, pId=100, key=node3, value=node3, childList=null)]]
    */
   System.out.println("tree:" + ToStringBuilder.reflectionToString(tree));
 }

private static Tree metaTree(List<Tree> treeList, Integer id) {
//此处getId getPId根据自己实际情况更改
   Tree treeConfig = treeList.stream().filter(tree -> tree.getId().equals(id)).collect(Collectors.toList()).get(0);
   Map<Integer, List<Tree>> collect = treeList.stream().filter(type -> type.getPId() != null).collect(Collectors.groupingBy(Tree::getPId));
   if (collect != null && collect.size() > 0) {
     recursion(collect, treeConfig);
   }
   return treeConfig;
 }

private static Tree recursion(Map<Integer, List<Tree>> maps, Tree tree) {
   if (tree.getChildList() == null) {
     if (maps.get(tree.getId()) != null) {
       tree.setChildList(maps.get(tree.getId()));
       maps.remove(tree.getId());
       if (maps.size() > 0) {
         recursion(maps, tree);
       }
     }
   } else {
     List<Tree> metaTypeList = tree.getChildList();
     if (metaTypeList != null && metaTypeList.size() > 0) {
       for (Tree meta : metaTypeList) {
         recursion(maps, meta);
       }
     }
   }
   return tree;
 }
}

来源：https://www.cnblogs.com/javashare/p/9814415.html