const treeData=[
{id:1,
name:'中国',
children:[
{id:11,name:'河南省',children:[{id:111,name:'南阳市',children:[{id:1111,name:'淅川县',children:null}]},{id:112,name:'郑州市',children:[{id:1121,name:'中牟县',children:null}]}]
},
{id:22,name:'广东省',children:[{id:221,name:'深圳市',children:[{id:2211,name:'宝安区',children:null},{id:2212,name:'福田区',children:null}]},{id:222,name:'广州市',children:[{id:2221,name:'天河区',children:null},{id:2222,name:'南沙区',children:null}]}]
}]}
]

const listData1=[
{id:1,name:'中国'},
{id:11,name:'河南省',pid:1},
{id:111,name:'南阳市',pid:11},
{id:1111,name:'淅川县',pid:111},
{id:112,name:'郑州市',pid:11},
{id:1121,name:'中牟县',pid:112},
{id:12,name:'广东省',pid:1},
{id:121,name:'深圳市',pid:12},
{id:1211,name:'宝安区',pid:121},
{id:1212,name:'福田区',pid:121},
{id:122,name:'广州市',pid:12},
{id:1221,name:'天河区',pid:122},
{id:1222,name:'南沙区',pid:122},
]

1、一维数组转树形结构

1.1、递归

//一维数组转树形结构
function listToTree(list){const tree=[]for(const node of list){//如果是父节点if(!node.pid){let p={...node}p.children=getChildren(p.id,list)tree.push(p)}}return tree
}
function getChildren(id,list){const children=[]//遍历数组的每一个元素，如果找到元素的父元素节点等于当前节点，则向children中添加该元素for(const node of list){if(node.pid===id){children.push(node)}}//遍历每一个孩子节点，递归寻找孩子的孩子节点，如果存在children，则向node的children中添加for(const node of children){const children=getChildren(node.id,list)if(children.length){node.children=children}}return children
}

demo：

console.log(listToTree(listData))

打印后的结果和treeData一致。

1.2、双层循环

// 双层循环：找出每一个元素的父节点，向每一个父节点增加children属性
function listToTree2(list){list.forEach(child=>{const pid=child.pid//遍历子节点，然后遍历每一个list节点，从中找出父节点，向父节点增加孩子元素if(pid){list.forEach(parent=>{if(parent.id===pid){//如果找到的父节点还没有孩子，会默认为空数组，否则为原来的子节点数组                    parent.children=parent.children||[]parent.children.push(child)}})}})
}

结果同1.1

1.3、使用map

//Map
function listToListMap(list){const [map,treeData]=[{},[]]//将map转为以id为key，索引为value的对象//将数组中每个元素的children初始化为空数组for(let i=0;i<list.length;i++){map[list[i].id]=i;list[i].children=[]}console.log(map)for(let i=0;i<list.length;i++){const node=list[i]//如果是子元素并且子元素的父节点存在在list中，则找到父元素并向父元素的children中加入子元素if(node.pid&&list[map[node.pid]]){list[map[node.pid]].children.push(node)}else{//说明不是子元素，是父元素直接放入treeData中treeData.push(node)}}return treeData
}

2、tree转list

树形结构转数组涉及到树的遍历，树的遍历分为深度遍历（前中后序）和广度遍历。

2.1、广度遍历

const treeData3=[
{id:1,
name:'中国',
children:[
{id:11,name:'河南省',children:[{id:111,name:'南阳市',pid:11,children:[{id:1111,name:'淅川县',children:null,pid:111,}]},{id:112,name:'郑州市',pid:11,children:[{id:1121,pid:112,name:'中牟县',children:null}]}]
},
{id:22,name:'广东省',children:[{id:221,pid:22,name:'深圳市',children:[{id:2211,pid:221,name:'宝安区',children:null},{id:2212,pid:221,name:'福田区',children:null}]},{id:222,pid:22,name:'广州市',children:[{id:2221,pid:222,name:'天河区',children:null},{id:2222,pid:222,name:'南沙区',children:null}]}]
}]}
]
//广度遍历(一层一层遍历)
function treeToList(treeData){const list=[]const queue=[...treeData]while(queue.length){//取队头元素         const node=queue.shift()const children=node.children//将队头元素的孩子依次放入队列中便于后序遍历if(children){queue.push(...children)}//放入元素节点list.push(node)}return list
}
console.log(treeToList(treeData3))

打印结果同1中的listData

2.2、深度遍历

//深度遍历(采用栈)
function treeToList2(tree){const list=[]const stack=[...tree]while(stack.length){const node=stack.pop()//如果出栈的节点有孩子元素,那么将出栈元素的所有孩子元素都放入栈中,便于后续遍历.const children=node.childrenif(children){stack.push(...children)}list.push(node)}return list
}

3、list转Map(map是以父节点id为key,所有子节点作为数组为value的数据结构)

const listData4=[
{id:1,name:'中国'},//0
{id:11,name:'河南省',pid:1},//1
{id:111,name:'南阳市',pid:11},//2
{id:1111,name:'淅川县',pid:111},//3
{id:112,name:'郑州市',pid:11},//4
{id:1121,name:'中牟县',pid:112},//5
{id:12,name:'广东省',pid:1},//6
{id:121,name:'深圳市',pid:12},//7
{id:1211,name:'宝安区',pid:121},//8
{id:1212,name:'福田区',pid:121},//9
{id:122,name:'广州市',pid:12},//10
{id:1221,name:'天河区',pid:122},//11
{id:1222,name:'南沙区',pid:122},//12
]
// 将列表数据转为以父节点id为key,子节点为value的map
function listToMap(list){const map={}list.forEach(item=>{for(let i=0;i<list.length;i++){//如果外层循环元素的id等于内层循环元素的pid,说明内层元素是外层元素的子元素,则向map的value中添加该内层元素.if(list[i].pid===item.id){map[item.id]=map[item.id]||[]map[item.id].push(list[i])}}})return map
}
console.log(listToMap(listData4))