Java list如何根據(jù)id獲取子節(jié)點

更新時間：2020年03月21日 11:29:20 作者：學無終

這篇文章主要介紹了Java list如何根據(jù)id獲取子節(jié)點,文中通過示例代碼介紹的非常詳細，對大家的學習或者工作具有一定的參考學習價值,需要的朋友可以參考下

工作中因業(yè)務需求，將數(shù)據(jù)庫中的樹狀結(jié)構(gòu)的數(shù)據(jù)根據(jù)父節(jié)點獲取所有的子節(jié)點

實現(xiàn)思路

1.獲取整個數(shù)據(jù)的list集合數(shù)據(jù)

2.將數(shù)據(jù)分組，java8 list有g(shù)roupby分組，java8之前的自己遍歷整理

3.分組后遞歸獲取子節(jié)點，有子節(jié)點的添加，沒有的設置子節(jié)點并刪除分組的數(shù)據(jù)，知道分組數(shù)據(jù)刪完

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) {
//此處getId getPId根據(jù)自己實際情況更改
    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;
  }
}

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