java 递归转非递归(树形转非树形) 2025615编辑

Heaven helps those who help themselves
资深码农+深耕理财=财富自由
欢迎关注

java 递归转非递归(树形转非树形)

Created by Marydon on 2022-05-25 17:37

1.情景展示

接着上一篇讲,在上一篇我们已经将数据通过递归转成了树形结构。

如何将树形结构转非树形结构?(仍然按照树形层级关系进行平铺显示)

如何将递归函数转成迭代函数?

查看代码
[{
name = 根节点,
childList = [{
name = 一级节点1,
childList = [{
name = 二级节点1.1,
childList = [],
id = 3,
parentId = 2
}, {
name = 二级节点1.2,
childList = [],
id = 4,
parentId = 2
}, {
name = 二级节点1.3,
childList = [],
id = 5,
parentId = 2
}
],
id = 2,
parentId = 1
}, {
name = 一级节点2,
childList = [{
name = 二级节点2.1,
childList = [{
name = 三级节点2.1.1,
childList = [],
id = 9,
parentId = 7
}, {
name = 三节点2.1.2,
childList = [],
id = 10,
parentId = 7
}
],
id = 7,
parentId = 6
}, {
name = 二级节点2.2,
childList = [],
id = 8,
parentId = 6
}
],
id = 6,
parentId = 1
}, {
name = 一级节点3,
childList = [{
name = 二级节点3.1,
childList = [{
name = 三级节点3.1.1,
childList = [{
name = 四级节点3.1.1.1,
childList = [{
name = 五级节点3.1.1.1.1,
childList = [],
id = 15,
parentId = 14
}
],
id = 14,
parentId = 13
}
],
id = 13,
parentId = 12
}
],
id = 12,
parentId = 11
}
],
id = 11,
parentId = 1
}
],
id = 1,
parentId = 0
}
]

2.具体分析

可以通过栈Stack来完成。

3.解决方案

代码实现

/*
* 将父节点下的子节点进行由嵌套改成平铺
* @description: 递归转迭代
* 按照层级关系,将嵌套改成平铺
* @param: root 嵌套数据
* @return: java.util.ArrayList<java.util.Map<java.lang.String,java.lang.Object>>
* 平铺数据
*/
public static ArrayList<Map<String, Object>> tileTraversal(Map<String, Object> root) {
ArrayList<Map<String, Object>> res = new ArrayList<>();
if (root == null) {
return res;
}
// 建立一个新的栈实例
Stack<Map<String, Object>> stack = new Stack<>();
// 把原始内容压入新建栈
stack.push(root);
while (!stack.empty()) {
// 出栈,将栈顶元素赋给temp
Map<String, Object> temp = stack.pop();
Map<String, Object> copyMap = new HashMap<>(temp.size());
// 复制当前节点
copyMap.putAll(temp);
// 移除map(该节点)当中的子节点
copyMap.remove("childList");
// 将此节点放到集合当中
res.add(copyMap);
// 当前节点,存在子节点
if (temp.get("childList") != null && ((List<Map<String, Object>>) temp.get("childList")).size() > 0) {
// ((List<Map<String, Object>>) temp.get("childList")).forEach(m -> stack.push(m));
List<Map<String, Object>> list = (List<Map<String, Object>>) temp.get("childList");
for (int i = list.size() - 1; i >= 0; i--) {
// 将子节点倒序进栈
stack.push(list.get(i));
}
}
}
return res;
}

调用

// mapList是树形结构(已经存在的数据)
List<Map<String, Object>> tileList = new ArrayList<>();
mapList.forEach(m -> tileList.addAll(tileTraversal(m)));
System.out.println(tileList);
[{name=根节点, id=1, parentId=0}, {name=一级节点1, id=2, parentId=1}, {name=二级节点1.1, id=3, parentId=2}, {name=二级节点1.2, id=4, parentId=2}, {name=二级节点1.3, id=5, parentId=2}, {name=一级节点2, id=6, parentId=1}, {name=二级节点2.1, id=7, parentId=6}, {name=三级节点2.1.1, id=9, parentId=7}, {name=三节点2.1.2, id=10, parentId=7}, {name=二级节点2.2, id=8, parentId=6}, {name=一级节点3, id=11, parentId=1}, {name=二级节点3.1, id=12, parentId=11}, {name=三级节点3.1.1, id=13, parentId=12}, {name=四级节点3.1.1.1, id=14, parentId=13}, {name=五级节点3.1.1.1.1, id=15, parentId=14}]

4.效果展示

转JSON

System.out.println(JSON.toJSON(tileList));
[{"name":"根节点","id":"1","parentId":"0"},{"name":"一级节点1","id":"2","parentId":"1"},{"name":"二级节点1.1","id":"3","parentId":"2"},{"name":"二级节点1.2","id":"4","parentId":"2"},{"name":"二级节点1.3","id":"5","parentId":"2"},{"name":"一级节点2","id":"6","parentId":"1"},{"name":"二级节点2.1","id":"7","parentId":"6"},{"name":"三级节点2.1.1","id":"9","parentId":"7"},{"name":"三节点2.1.2","id":"10","parentId":"7"},{"name":"二级节点2.2","id":"8","parentId":"6"},{"name":"一级节点3","id":"11","parentId":"1"},{"name":"二级节点3.1","id":"12","parentId":"11"},{"name":"三级节点3.1.1","id":"13","parentId":"12"},{"name":"四级节点3.1.1.1","id":"14","parentId":"13"},{"name":"五级节点3.1.1.1.1","id":"15","parentId":"14"}]

格式化

查看代码
[{
"name": "根节点",
"id": "1",
"parentId": "0"
},
{
"name": "一级节点1",
"id": "2",
"parentId": "1"
},
{
"name": "二级节点1.1",
"id": "3",
"parentId": "2"
},
{
"name": "二级节点1.2",
"id": "4",
"parentId": "2"
},
{
"name": "二级节点1.3",
"id": "5",
"parentId": "2"
},
{
"name": "一级节点2",
"id": "6",
"parentId": "1"
},
{
"name": "二级节点2.1",
"id": "7",
"parentId": "6"
},
{
"name": "三级节点2.1.1",
"id": "9",
"parentId": "7"
},
{
"name": "三节点2.1.2",
"id": "10",
"parentId": "7"
},
{
"name": "二级节点2.2",
"id": "8",
"parentId": "6"
},
{
"name": "一级节点3",
"id": "11",
"parentId": "1"
},
{
"name": "二级节点3.1",
"id": "12",
"parentId": "11"
},
{
"name": "三级节点3.1.1",
"id": "13",
"parentId": "12"
},
{
"name": "四级节点3.1.1.1",
"id": "14",
"parentId": "13"
},
{
"name": "五级节点3.1.1.1.1",
"id": "15",
"parentId": "14"
}]

说明:

事实上,使用SQL进行递归查询,返回的就是这种格式的数据,即:按照层级关系排好序展示出来;

如果前端需要的是这种排好序的数据,直接返回即可;

如果需要返回树形结构的话,见文末推荐。 

写在最后

  哪位大佬如若发现文章存在纰漏之处或需要补充更多内容,欢迎留言!!!

 相关推荐:

与君共勉:最实用的自律是攒钱,最养眼的自律是健身,最健康的自律是早睡,最改变气质的自律是看书,最好的自律是经济独立 。

您的一个点赞,一句留言,一次打赏,就是博主创作的动力源泉!

↓↓↓↓↓↓写的不错,对你有帮助?赏博主一口饭吧↓↓↓↓↓↓

posted @   Marydon  阅读(615)  评论(0编辑  收藏  举报
相关博文:
阅读排行:
· 全程不用写代码,我用AI程序员写了一个飞机大战
· DeepSeek 开源周回顾「GitHub 热点速览」
· 记一次.NET内存居高不下排查解决与启示
· MongoDB 8.0这个新功能碉堡了,比商业数据库还牛
· .NET10 - 预览版1新功能体验(一)
历史上的今天:
2021-05-25 windows/linux 同时运行两个以上tomcat
2018-05-25 chrome 此网页正试图从未经验证的来源加载脚本
2018-05-25 css 设置英文字母大小写转换(text-transform)
2017-05-25 javascript 自定义Map
2017-05-25 javascript Array(数组)
点击右上角即可分享
微信分享提示