IT俱乐部 Java Java实现树形List与扁平List互转的示例代码

Java实现树形List与扁平List互转的示例代码

发布: 2023年 5月 12日 188阅读

背景：在平时的开发中，我们时常会遇到下列场景

  • 公司的组织架构的数据存储与展示
  • 文件夹层级的数据存储与展示
  • 评论系统中，父评论与诸多子评论的数据存储与展示
  • ……

对于这种有层级的结构化数据，就像是一棵树一样。在关系型数据库中，通常将一个个的节点信息存储到表中，通过一个字段（例如，pid），指向其父节点。而在数据展示的时候，我们又希望它是呈现层级的，例如：

id  name        pid
1   总公司       -1
2   上海分公司    1
3   浙江分公司    1
4   闵行区分公司  2
5   嘉兴分公司    3
{
    "id": 1,
    "name": "总公司",
    "pid": -1,
    "branch":
    [
        {
            "id": 2,
            "name": "上海分公司",
            "pid": 1,
            "branch":
            [
                {
                    "id": 4,
                    "name": "闵行区分公司",
                    "pid": 2,
                    "branch":
                    []
                }
            ]
        },
        {
            "id": 3,
            "name": "浙江分公司",
            "pid": 1,
            "branch":
            [
                {
                    "id": 5,
                    "name": "嘉兴分公司",
                    "pid": 3,
                    "branch":
                    []
                }
            ]
        }
    ]
}

所以，本文的主要内容就是提供几种方案，实现这两类数据的转换方式。

存储树的表结构

如何在一张数据库表中表示一颗树结构中的所有节点信息，这里有一个示例：

DROP TABLE IF EXISTS zj_city;
CREATE TABLE zj_city (
    id BIGINT NOT NULL AUTO_INCREMENT,
    name VARCHAR(50) COMMENT '节点名称',
    pid int NOT NULL COMMENT '父节点',

    create_time DATETIME DEFAULT now() COMMENT '创建时间: 年-月-日 时:分:秒',
    update_time DATETIME DEFAULT now() ON UPDATE now() COMMENT '更新时间',
    is_deleted BIT DEFAULT 0 COMMENT '是否删除：0-false-未删除;1-true-已删除',
    PRIMARY KEY (id)
)COMMENT '浙江城市';

INSERT INTO zj_city(name, pid) VALUES
('浙江省',0),
('金华市',1),
('嘉兴市',1),
('杭州市',1),
('宁波市',1);

INSERT INTO zj_city(name, pid) VALUES
('下城区',4),
('钱塘区',4),
('西湖区',4),
('上城区',4);

INSERT INTO zj_city(name, pid) VALUES
('南湖区',3),
('秀洲区',3),
('桐乡市',3),
('平湖市',3),
('海宁市',3);

INSERT INTO zj_city(name, pid) VALUES
('梧桐街道',12),
('凤鸣街道',12),
('龙翔街道',12),
('崇福镇',12),
('乌镇镇',12),
('高桥镇',12),
('河山镇',12),
('濮院镇',12);

SELECT * from zj_city;

扁平List转树形List

应用场景

  • 公司组织结构
  • 省市级
  • 评论系统中，父评论与诸多子评论
  • 其他层级展示的数据

双层for

主要思想：外层循环-找父节点；内层循环-找子节点；因为每个元素都会找一遍，所有最终得到完整的树

import com.alibaba.fastjson.JSON;
import com.ks.boot.entity.CityEntity;
import org.junit.jupiter.api.BeforeEach;
import org.junit.jupiter.api.Test;
import java.util.ArrayList;
import java.util.List;
public class TreeListDemo {
    List cityEntities;
    /**
     * id  name  pid
     * 1   浙江   0
     * 2   杭州   1
     * 3   嘉兴   1
     * 4   南湖   3
     * 5   桐乡   3
     * 6   余杭   2
     * 7   西湖   2
     * 8   云南   0
     * 9   昆明   8
     * 10  昭通   8
     */
    @BeforeEach
    public void init() {
        cityEntities = JSON.parseArray("[{"id":1,"name":"浙江","pid":0},n" +
                "{"id":2,"name":"杭州","pid":1},n" +
                "{"id":3,"name":"嘉兴","pid":1},n" +
                "{"id":4,"name":"南湖","pid":3},n" +
                "{"id":5,"name":"桐乡","pid":3},n" +
                "{"id":6,"name":"余杭","pid":2},n" +
                "{"id":7,"name":"西湖","pid":2},n" +
                "{"id":8,"name":"云南","pid":0},n" +
                "{"id":9,"name":"昆明","pid":8},n" +
                "{"id":10,"name":"昭通","pid":8}]", CityEntity.class);
    }
    @Test
    public void testList2Tree() {
        List resultTree = list2Tree(this.cityEntities);
        System.out.println(JSON.toJSONString(resultTree));
    }
    /**
     * 双层for，List转Tree
     * 主要思想：外层循环-找父节点；内层循环-找子节点；因为每个元素都会找一遍，所有最终得到完整的树
     * 时间复杂度：N^2；空间复杂度：N
     */
    public List list2Tree(List cityEntities) {
        List resultCities = new ArrayList();
        for (CityEntity city : cityEntities) {
            if(0 == city.getPid()) { //根节点、顶级节点，直接放入最终返回结果的List
                resultCities.add(city);
            }
            for (CityEntity curCity : cityEntities) { //根据当前city找它的子节点
                if(city.getId().equals(curCity.getPid())) {
                    if(city.getSubCityList() == null) { //还没有任何子节点，new一个空的放进去
                        city.setSubCityList(new ArrayList());
                    }
                    city.getSubCityList().add(curCity);
                }
            }
        }
        return resultCities;
    }
}
public class CityEntity {
    private Long id;
    private String name;
    private Long pid;
    private List subCityList;
    getter/setter
}

递归

主要思想：获取所有根节点、顶级节点，再从List中查找根节点的子节点；

import com.alibaba.fastjson.JSON;
import com.ks.boot.entity.CityEntity;
import org.junit.jupiter.api.BeforeEach;
import org.junit.jupiter.api.Test;
import java.util.ArrayList;
import java.util.List;
public class TreeListDemo {
    List cityEntities;
    /**
     * id  name  pid
     * 1   浙江   0
     * 2   杭州   1
     * 3   嘉兴   1
     * 4   南湖   3
     * 5   桐乡   3
     * 6   余杭   2
     * 7   西湖   2
     * 8   云南   0
     * 9   昆明   8
     * 10  昭通   8
     */
    @BeforeEach
    public void init() {
        cityEntities = JSON.parseArray("[{"id":1,"name":"浙江","pid":0},n" +
                "{"id":2,"name":"杭州","pid":1},n" +
                "{"id":3,"name":"嘉兴","pid":1},n" +
                "{"id":4,"name":"南湖","pid":3},n" +
                "{"id":5,"name":"桐乡","pid":3},n" +
                "{"id":6,"name":"余杭","pid":2},n" +
                "{"id":7,"name":"西湖","pid":2},n" +
                "{"id":8,"name":"云南","pid":0},n" +
                "{"id":9,"name":"昆明","pid":8},n" +
                "{"id":10,"name":"昭通","pid":8}]", CityEntity.class);
    }
    /**
     * 递归，List转Tree
     * 主要思想：获取所有根节点、顶级节点，再从List中查找根节点的子节点；
     * 时间复杂度：N*(1+N)/2，O(N^2)，因为递归方法中，最好是List的第一元素就是要找的子节点，最坏是
     * List的最后一个元素是子节点
     */
    @Test
    public void testList2Tree() {
        List resultCities = new ArrayList();
        for (CityEntity city : cityEntities) {
            if(0 == city.getPid()) { //获取所有根节点、顶级节点，再根据根节点进行递归
                CityEntity topCity = findChild(cityEntities, city); //此时的topCity已经包含它的所有子节点
                resultCities.add(topCity);
            }
        }
        System.out.println(JSON.toJSONString(resultCities));
    }
    /**
     *
     * @param cityEntities 在那个里面找
     * @param curCity 找谁的子节点
     * @return curCity的子节点
     */
    public CityEntity findChild(List cityEntities, CityEntity curCity) {
        for (CityEntity city : cityEntities) {
            if(curCity.getId().equals(city.getPid())) {
                if(null == curCity.getSubCityList()) {
                    curCity.setSubCityList(new ArrayList());
                }
                CityEntity subChild = findChild(cityEntities, city); //每次递归，都从头开始查找有没有city的子节点
                curCity.getSubCityList().add(subChild);
            }
        }
        return curCity;
    }
}
public class CityEntity {
    private Long id;
    private String name;
    private Long pid;
    private List subCityList;
    getter/setter
}

转换为Map

主要思想：

  • 在双层for的解法中，由于内层for也需要遍历以便List，造成时间复杂度上身为平方级
  • 如果内层for不需要遍历完整的List，而是事先预处理到Map中，寻找时直接从Map中获取，则时间复杂度降为LogN
import com.alibaba.fastjson2.JSON;
import org.junit.jupiter.api.BeforeEach;
import org.junit.jupiter.api.Test;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
public class TreeListDemo {
    List cityEntities;
    /**
     * id  name  pid
     * 1   浙江   0
     * 2   杭州   1
     * 3   嘉兴   1
     * 4   南湖   3
     * 5   桐乡   3
     * 6   余杭   2
     * 7   西湖   2
     * 8   云南   0
     * 9   昆明   8
     * 10  昭通   8
     */
    @BeforeEach
    public void init() {
        cityEntities = JSON.parseArray("[{"id":1,"name":"浙江","pid":0},n" +
                "{"id":2,"name":"杭州","pid":1},n" +
                "{"id":3,"name":"嘉兴","pid":1},n" +
                "{"id":4,"name":"南湖","pid":3},n" +
                "{"id":5,"name":"桐乡","pid":3},n" +
                "{"id":6,"name":"余杭","pid":2},n" +
                "{"id":7,"name":"西湖","pid":2},n" +
                "{"id":8,"name":"云南","pid":0},n" +
                "{"id":9,"name":"昆明","pid":8},n" +
                "{"id":10,"name":"昭通","pid":8}]", CityEntity.class);
    }
    /**
     * 在双层for的解法中，由于内层for也需要遍历以便List，造成时间复杂度上身为平方级
     * 如果内层for不需要遍历完整的List，而是事先预处理到Map中，寻找时直接从Map中获取，则时间复杂度降为LogN
     */
    @Test
    public void list2tree() {
        //收集每个PID下的节点为Map
        Map> cityMapByPid = cityEntities.stream().collect(Collectors.groupingBy(CityEntity::getPid));
        List resultCityList = new ArrayList();
        for (CityEntity city : cityEntities) {
            if(0 == city.getPid()) { //根节点、顶点
                resultCityList.add(city);
            }
            List citiesByPid = cityMapByPid.get(city.getId());
            if(null != citiesByPid && citiesByPid.size() > 0) { //有子节点
                if(null == city.getSubCityList()) {
                    city.setSubCityList(new ArrayList());
                }
                city.getSubCityList().addAll(citiesByPid); //塞入
            }
        }
        System.out.println(JSON.toJSONString(resultCityList));
    }
    /**
     * 简化写法：在收集到Map时，对于没有子节点的节点，创建一个空的塞入到Map中
     */
    @Test
    public void list2tree4Simple() {
        List resultCityList = new ArrayList();
        //保存每个PID下的子节点
        Map> cityMapByPid = new HashMap();
        for (CityEntity city : cityEntities) { //收集每个PID下的子节点
            //获取当前PID对应的子节点List，如果没有则创建一个空的List塞入
            //这个设计得很巧妙
            List children = cityMapByPid.getOrDefault(city.getPid(), new ArrayList());
            children.add(city); //插入当前元素
            cityMapByPid.put(city.getPid(), children);
        }
        for (CityEntity city : cityEntities) {
            if(0 == city.getPid()) { //根节点、顶点
                resultCityList.add(city);
            }
            city.setSubCityList(cityMapByPid.get(city.getId()));
        }
        System.out.println(JSON.toJSONString(resultCityList));
    }
}

栈

主要思想

收集根节点、顶级节点，存入resultList，并且压栈

循环出栈，栈元素Cur

  • 找Cur的所有子元素为child
  • 如果child不为空，则再压入栈中。这一步的目的是，再一次找child的子元素
import com.alibaba.fastjson2.JSON;
import org.junit.jupiter.api.BeforeEach;
import org.junit.jupiter.api.Test;
import java.util.*;
import java.util.stream.Collectors;
public class TreeListDemo {
    List cityEntities;
    /**
     * id  name  pid
     * 1   浙江   0
     * 2   杭州   1
     * 3   嘉兴   1
     * 4   南湖   3
     * 5   桐乡   3
     * 6   余杭   2
     * 7   西湖   2
     * 8   云南   0
     * 9   昆明   8
     * 10  昭通   8
     */
    @BeforeEach
    public void init() {
        cityEntities = JSON.parseArray("[{"id":1,"name":"浙江","pid":0},n" +
                "{"id":2,"name":"杭州","pid":1},n" +
                "{"id":3,"name":"嘉兴","pid":1},n" +
                "{"id":4,"name":"南湖","pid":3},n" +
                "{"id":5,"name":"桐乡","pid":3},n" +
                "{"id":6,"name":"余杭","pid":2},n" +
                "{"id":7,"name":"西湖","pid":2},n" +
                "{"id":8,"name":"云南","pid":0},n" +
                "{"id":9,"name":"昆明","pid":8},n" +
                "{"id":10,"name":"昭通","pid":8}]", CityEntity.class);
    }
    /**
     * 主要思想：
     *  收集根节点、顶级节点，存入resultList，并且压栈
     *  循环出栈，栈元素Cur
     *      找Cur的所有子元素为child
     *      如果child不为空，则再压入栈中。这一步的目的是，再一次找child的子元素
     * 时间复杂度：N(过滤出所有跟节点) + 常数(出栈) * N(遍历List找当前节点的子元素)
     */
    @Test
    public void list2tree() {
        List resultCityList = new ArrayList();
        Stack stack = new Stack();
        resultCityList = cityEntities.stream().filter(ele -> 0 == ele.getPid()).collect(Collectors.toList());
        stack.addAll(resultCityList); //根节点、顶点入栈
        while(!stack.isEmpty()) {
            CityEntity curCity = stack.pop();
            List child = cityEntities.stream().filter(ele -> curCity.getId().equals(ele.getPid())).collect(Collectors.toList());
            if(!child.isEmpty()) { //这一步处理的原因是：当没有子元素，不显示该个字段。流处理后没有元素只会返回空List，不会返回null
                curCity.setSubCityList(child);
            }
            if(!child.isEmpty()) {
                stack.addAll(child);
            }
        }
        System.out.println(JSON.toJSONString(resultCityList));
    }
}

树形List转扁平List

递归

主要思想：遍历树节点，一个树节点如果有子树，则再次递归此子树，直到没有子树为止

import com.alibaba.fastjson2.JSON;
import org.junit.jupiter.api.BeforeEach;
import org.junit.jupiter.api.Test;
import java.util.ArrayList;
import java.util.List;
/**
 * id  name  pid
 * 1   浙江   0
 * 2   杭州   1
 * 3   嘉兴   1
 * 4   南湖   3
 * 5   桐乡   3
 * 6   余杭   2
 * 7   西湖   2
 * 8   云南   0
 * 9   昆明   8
 * 10  昭通   8
 */
public class ListTreeDemo {
    List treeList;
    @BeforeEach
    public void init() {
        treeList = JSON.parseArray("[{"id":1,"name":"浙江","pid":0,"subCityList":[" +
                "{"id":2,"name":"杭州","pid":1,"subCityList":[{"id":6,"name":"余杭","pid":2},{"id":7,"name":"西湖","pid":2}]}," +
                "{"id":3,"name":"嘉兴","pid":1,"subCityList":[{"id":4,"name":"南湖","pid":3},{"id":5,"name":"桐乡","pid":3}]}]}," +
                "{"id":8,"name":"云南","pid":0,"subCityList":[{"id":9,"name":"昆明","pid":8},{"id":10,"name":"昭通","pid":8}]}]", CityEntity.class);
    }
    @Test
    public void tree2list() {
        List resList = new ArrayList();
        //这一层for的目的是：遍历根节点
        for (CityEntity city : treeList) {
            reversion(city,resList);
        }
        System.out.println(JSON.toJSONString(resList));
    }
    public void reversion(CityEntity curNode, List resList) {
        resList.add(beanCopy(curNode));
        List subCityList = curNode.getSubCityList();
        if(subCityList != null && !subCityList.isEmpty()) {
            for (CityEntity city : subCityList) { //递归寻找子节点的子节点们
                reversion(city, resList);
            }
        }
        //递归的出口就是subCityList为null或者empty
    }
    private CityEntity beanCopy(CityEntity source) {
        CityEntity res = new CityEntity();
        res.setId(source.getId());
        res.setName(source.getName());
        res.setPid(source.getPid());
        return res;
    }
}

栈

主要思想

依次遍历树形List，当前节点为Cur

  • 将Cur收集到某个存储结果的List
  • 如果Cur有子树，压入某个栈中

依次弹出栈元素，当前弹出的元素为StackSubTree

  • 如果StackSubTree还有子树，继续压栈
  • 如果StackSubTree没有子树，则放入结果List
import com.alibaba.fastjson2.JSON;
import org.junit.jupiter.api.BeforeEach;
import org.junit.jupiter.api.Test;

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

/**
 * id  name  pid
 * 1   浙江   0
 * 2   杭州   1
 * 3   嘉兴   1
 * 4   南湖   3
 * 5   桐乡   3
 * 6   余杭   2
 * 7   西湖   2
 * 8   云南   0
 * 9   昆明   8
 * 10  昭通   8
 */
public class ListTreeDemo {
    List treeList;

    @BeforeEach
    public void init() {
        treeList = JSON.parseArray("[{"id":1,"name":"浙江","pid":0,"subCityList":[" +
                "{"id":2,"name":"杭州","pid":1,"subCityList":[{"id":6,"name":"余杭","pid":2},{"id":7,"name":"西湖","pid":2}]}," +
                "{"id":3,"name":"嘉兴","pid":1,"subCityList":[{"id":4,"name":"南湖","pid":3},{"id":5,"name":"桐乡","pid":3}]}]}," +
                "{"id":8,"name":"云南","pid":0,"subCityList":[{"id":9,"name":"昆明","pid":8},{"id":10,"name":"昭通","pid":8}]}]", CityEntity.class);
    }

    /**
     * 1. 依次遍历树形List，当前节点为Cur
     *   a) 将Cur收集到某个存储结果的List
     *   b) 如果Cur有子树，压入某个栈中
     * 2. 依次弹出栈元素，当前弹出的元素为StackSubTree
     *   a) 如果StackSubTree还有子树，继续压栈
     *   b) 如果StackSubTree没有子树，则放入结果List
     */
    @Test
    public void tree2list() {
        List resList = new ArrayList();

        Stack> stack = new Stack();

        for (CityEntity curCity : treeList) {
            resList.add(beanCopy(curCity));
            if (curCity.getSubCityList() != null && !curCity.getSubCityList().isEmpty()) {
                stack.push(curCity.getSubCityList());
            }
        }

        while (!stack.isEmpty()) {
            List subTree = stack.pop();
            for (CityEntity city : subTree) {
                if (city.getSubCityList() != null && !city.getSubCityList().isEmpty()) {
                    stack.push(city.getSubCityList());
                } else {
                    resList.add(beanCopy(city));
                }
            }
        }

        System.out.println(JSON.toJSONString(resList));
    }

    private CityEntity beanCopy(CityEntity source) {
        CityEntity res = new CityEntity();
        res.setId(source.getId());
        res.setName(source.getName());
        res.setPid(source.getPid());
        return res;
    }
}

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