递归是什么？
link:baijiahao.baidu.com/s?id=173674…
递归，就是在运行的过程中不断地调用自己。递归有两个过程，简单地说一个是递的过程，一个是归的过程。
获取树形数据层级路径问题
数据源
listData: [
{
id: "1",
name: "test1",
children: [
{
id: "1-1",
name: "test1-1",
children: [
{
id: "1-1-1",
name: "test1-1-1",
children: [],
},
],
},
{
id: "1-2",
name: "test1-2",
children: [
{
id: "1-2-1",
name: "test1-2-1",
children: [],
},
],
},
],
},
{
id: "2",
name: "test2",
children: [],
},
{
id: "3",
name: "test3",
children: [],
},
{
id: "4",
name: "test4",
children: [
{
id: "4-1",
name: "test4-1",
children: [],
},
{
id: "4-2",
name: "test4-2",
children: [
{
id: "4-2-1",
name: "test4-2-1",
children: [],
},
{
id: "4-2-2",
name: "test4-2-2",
children: [
{
id: "4-2-2-1",
name: "test4-2-2-1",
children: [],
},
],
},
],
},
],
},
],
函数
getPath(list, arr = []) {
for (let i in list) {
let el = list[i];
arr.push(el.id);
if (el.id === "4-2-2-1") {
return { id: el.id };
} else if (el.children && el.children.length) {
let ars = this.getPath(el.children, arr);
if (ars) {
return ars;
}
}
arr.pop();
}
},
// this.getPath(this.listData)
扁平数据转树形结构
数据源
listData: [
{
id: 1,
pid: 0,
name: "test1",
},
{
id: 2,
pid: 1,
name: "test2",
},
{
id: 3,
pid: 1,
name: "test3",
},
{
id: 4,
pid: 2,
name: "test4",
},
{
id: 5,
pid: 2,
name: "test5",
},
{
id: 6,
pid: 5,
name: "test6",
},
],
实现函数
getTree(list) {
let keys = {};
let tree = [];
list.forEach((el, index) => {
el.children = [];
let { id } = el;
keys[id] = el;
});
for (let i in list) {
let el = list[i];
let newItem = keys[el.id];
if (el.pid in keys) {
keys[el.pid]["children"].push(el);
} else {
tree.push(newItem);
}
}
return tree;
},
树形结构转扁平数据
数据源
listDataCopy: [
{
id: 1,
pid: 0,
name: "test1",
children: [
{
id: 2,
pid: 1,
name: "test2",
children: [
{
id: 4,
pid: 2,
name: "test4",
children: [],
},
{
id: 5,
pid: 2,
name: "test5",
children: [
{
id: 6,
pid: 5,
name: "test6",
children: [],
},
],
},
],
},
{
id: 3,
pid: 1,
name: "test3",
children: [],
},
],
},
],
方法
daping(list, arr = []) {
list.forEach((el) => {
arr.push(el);
if (el.children && el.children.length) {
this.daping(el.children, arr);
}
});
return arr;
},
