java将list转为树形结构的方法

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); } }

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); } }