java将list转为树形结构的方法

目录

• 1、通过转化成json封装数据
• 2、通过java8 stream转换

1、通过转化成json封装数据

原始数据如下

[
    {
        "name":"甘肃省",
        "pid":0,
        "id":1
    },
    {
        "name":"天水市",
        "pid":1,
        "id":2
    },
    {
        "name":"秦州区",
        "pid":2,
        "id":3
    },
    {
        "name":"北京市",
        "pid":0,
        "id":4
    },
    {
        "name":"昌平区",
        "pid":4,
        "id":5
    }
]

现需要是使用java将以上数据转为树形结构，转化后下的结构如下

[
    {
        "children":[
            {
                "children":[
                    {
                        "name":"秦州区",
                        "pid":2,
                        "id":3
                    }
                ],
                "name":"天水市",
                "pid":1,
                "id":2
            }
        ],
        "name":"甘肃省",
        "pid":0,
        "id":1
    },
    {
        "children":[
            {
                "name":"昌平区",
                "pid":4,
                "id":5
            }
        ],
        "name":"北京市",
        "pid":0,
        "id":4
    }
]

代码如下

/**

- listToTree
- <p>方法说明<p>
- 将JSONArray数组转为树状结构
- @param arr 需要转化的数据
- @param id 数据唯一的标识键值
- @param pid 父id唯一标识键值
- @param child 子节点键值
- @return JSONArray
*/
public static JSONArray listToTree(JSONArray arr,String id,String pid,String child){
   JSONArray r = new JSONArray();
   JSONObject hash = new JSONObject();
   //将数组转为Object的形式，key为数组中的id
   for(int i=0;i<arr.size();i++){
  JSONObject json = (JSONObject) arr.get(i);
  hash.put(json.getString(id), json);
   }
   //遍历结果集
   for(int j=0;j<arr.size();j++){
  //单条记录
  JSONObject aVal = (JSONObject) arr.get(j);
  //在hash中取出key为单条记录中pid的值
  JSONObject hashVP = (JSONObject) hash.get(aVal.get(pid).toString());
  //如果记录的pid存在，则说明它有父节点，将她添加到孩子节点的集合中
  if(hashVP!=null){
     //检查是否有child属性
     if(hashVP.get(child)!=null){
        JSONArray ch = (JSONArray) hashVP.get(child);
        ch.add(aVal);
        hashVP.put(child, ch);
     }else{
        JSONArray ch = new JSONArray();
        ch.add(aVal);
        hashVP.put(child, ch);
     }
  }else{
     r.add(aVal);
  }
   }
   return r;
}
测试代码如下
public static void main(String[] args){
   List<Map<String,Object>> data = new ArrayList<>();
   Map<String,Object> map = new HashMap<>();
   map.put("id",1);
   map.put("pid",0);
   map.put("name","甘肃省");
   data.add(map);
   Map<String,Object> map2 = new HashMap<>();
   map2.put("id",2);
   map2.put("pid",1);
   map2.put("name","天水市");
   data.add(map2);
   Map<String,Object> map3 = new HashMap<>();
   map3.put("id",3);
   map3.put("pid",2);
   map3.put("name","秦州区");
   data.add(map3);
   Map<String,Object> map4 = new HashMap<>();
   map4.put("id",4);
   map4.put("pid",0);
   map4.put("name","北京市");
   data.add(map4);
   Map<String,Object> map5 = new HashMap<>();
   map5.put("id",5);
   map5.put("pid",4);
   map5.put("name","昌平区");
   data.add(map5);
   System.out.println(JSON.toJSONString(data));
   JSONArray result = listToTree(JSONArray.parseArray(JSON.toJSONString(data)),"id","pid","children");
   System.out.println(JSON.toJSONString(result));
}

2、通过java8 stream转换

我在网上找了很多方法，自己写的这个思路清晰，代码量少，希望能找到志同道合的朋友，看看还有没有优化的地方。

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
@Data
@AllArgsConstructor
@NoArgsConstructor
public class ZhField {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private String id;

    /**
     * 上级领域id
     */
    @Column(name = "parent_id")
    private String parentId;

    /**
     * 领域名称
     */
    private String name;

    /**
     * 排序
     */
    private Integer sort;

}

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;

import java.io.Serializable;
import java.util.List;
import java.util.Map;

@Data
@AllArgsConstructor
@NoArgsConstructor
public class TreeMenuNode implements Serializable {
	private String id;
	private String parentId;
	private String name;
	private Integer sort;
	private List<TreeMenuNode> children;
	private Boolean isAble;

	/**20180929zhw添加 树的额外属性(至少含有父节点ID:"parentId")**/
	private Map<String,Object> attributes;
}

package com.egaoqi.service.impl.company;

import com.egaoqi.entity.ZhField;
import com.egaoqi.util.TreeMenuNode;

import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

/**
 * @author zhizhao
 * @description
 * @create 2018-11-14 9:07
 */
public class toModel {

    private static void forEach(Map<String, List<TreeMenuNode>> collect, TreeMenuNode treeMenuNode) {
        List<TreeMenuNode> treeMenuNodes = collect.get(treeMenuNode.getId());
        if(collect.get(treeMenuNode.getId())!=null){
            //排序
            treeMenuNodes.sort((u1, u2) -> u1.getSort().compareTo(u2.getSort()));
            treeMenuNodes.stream().sorted(Comparator.comparing(TreeMenuNode::getSort)).collect(Collectors.toList());
            treeMenuNode.setChildren(treeMenuNodes);
            treeMenuNode.getChildren().forEach(t->{
                forEach(collect,t);
            });
        }
    }

    public static void main(String[] args) {
        List<ZhField> zhFields = new ArrayList<>();
        List<TreeMenuNode> treeNodeList = new ArrayList<>();
        //转换数据,这个是前端需要的格式。
        zhFields.forEach(t->{
            TreeMenuNode treeMenuNode = new TreeMenuNode();
            treeMenuNode.setId(t.getId());
            treeMenuNode.setParentId(t.getParentId());
            treeMenuNode.setName(t.getName());
            treeMenuNode.setSort(t.getSort());
            treeNodeList.add(treeMenuNode);
        });
        //分组
        Map<String, List<TreeMenuNode>> collect = treeNodeList.stream().collect(Collectors.groupingBy(TreeMenuNode::getParentId));
        //树形结构 肯定有一个根部，我的这个根部的就是parentId.euqal("0"),而且只有一个就get（"0"）
        TreeMenuNode treeMenuNode = collect.get("0").get(0);
        //拼接数据
        forEach(collect, treeMenuNode);
    }
}