前提：数据库中查出来每一条数据构成链表，需要我们转换成树结构，id，pid（父节点的id） name
主要代码：public Node getRoot(List<Node> list)
0为根节点 1-9为次根节点 1.1-1.9 2.1-2.9 … 为次次根节点 看下方测试结果注意只有0的pid为-1
@RestController public class AAA { @RequestMapping("/project/test") public Result get123(){ Test test = new Test(); List<Node> list = new ArrayList<>(); list.add(new Node("0","-1","根节点")); list.add(new Node("9","0","9")); list.add(new Node("9.1","9","9.1")); list.add(new Node("1","0","1")); list.add(new Node("1.1","1","1.1")); list.add(new Node("2","0","2")); list.add(new Node("2.1","2","2.1")); list.add(new Node("3","0","3")); list.add(new Node("3.1","3","3.1")); list.add(new Node("3.1.2","3.1","3.1.2")); Node root = test.getRoot(list); Result<Object> result = new Result<>(); System.out.println(root); result.setResult(root); return result; } } class Node{ String id; String pid; String name; List<Node> children; public Node(String id,String pid,String name){ this.id=id; this.pid = pid; this.name = name; this.children = new ArrayList<>(); } public List<Node> getChildren() { return children; } public void setChildren(List<Node> children) { this.children = children; } public void setId(String id){ this.id = id; } public void setPid(String pid) { this.pid = pid; } public void setName(String name){ this.name = name; } public String getId() { return id; } public String getName() { return name; } public String getPid() { return pid; } } class Test{ public Node getRoot(List<Node> list){ Node root=null; Map<String,Node> map= new HashMap<String,Node>(); for(Node n:list){ String id = n.getId(); Node node = map.get(id); if(node==null){ node = n; map.put(id,n); }else{ node = map.get(id); node.setPid(n.getPid()); node.setName(n.getName()); } String pid = node.getPid(); if(pid.equals("-1")) { //根结点的判断 root = node; continue; } Node node1 = map.get(pid); if(node1==null){ node1 = new Node(pid,"-1","头节点"); map.put(pid,node1); } node1.getChildren().add(node); } return root; } }
测试结果
{ "success": true, "message": "操作成功！", "code": 0, "result": { "id": "0", "pid": "-1", "name": "根节点", "children": [ { "id": "9", "pid": "0", "name": "9", "children": [ { "id": "9.1", "pid": "9", "name": "9.1", "children": [ ] } ] }, { "id": "1", "pid": "0", "name": "1", "children": [ { "id": "1.1", "pid": "1", "name": "1.1", "children": [ ] } ] }, { "id": "2", "pid": "0", "name": "2", "children": [ { "id": "2.1", "pid": "2", "name": "2.1", "children": [ ] } ] }, { "id": "3", "pid": "0", "name": "3", "children": [ { "id": "3.1", "pid": "3", "name": "3.1", "children": [ { "id": "3.1.2", "pid": "3.1", "name": "3.1.2", "children": [ ] } ] } ] } ] }, "timestamp": 1651161878529 }
解释：
- 每次先查看map中是不是有id对应的key，如果有的话，说明之前添加过但是添加的对象只有id和children，这时候就需要把对应value中的对象的其他变量赋值，这个为什么只有id一会看到下边就知道了。
如果没有对应的key是id的话，添加key为id ，value为Node对象。 - 1中添加完id对应的Node之后，如果当前id节点的pid节点已存在，则把当前id节点添加到pid节点children数组中。如果不存在，需要先创建一个pid的新节点，并把新节点的id=pid，新节点添加到map中，并把1中的Node节点添加到新结点的children中（知道1中那个为什么只有id和children了吧）。
