/*
let arr = [
{ id: 1, name: '部门1', pid: 0 },
{ id: 2, name: '部门2', pid: 1 },
{ id: 3, name: '部门3', pid: 1 },
{ id: 4, name: '部门4', pid: 3 },
{ id: 5, name: '部门5', pid: 4 },
{ id: 6, name: '部门6', pid: 4666 },
]
输出结果:
[
{
"id": 1,
"name": "部门1",
"pid": 0,
"children": [
{
"id": 2,
"name": "部门2",
"pid": 1,
"children": []
},
{
"id": 3,
"name": "部门3",
"pid": 1,
"children": [
// 结果 ,,,
]
}
]
},
{
id: 6,
name: '部门6',
pid: 4666
}
]
*/
/*
方案1 递归方式
先找出所有根节点，再找出所有根节点的子节点
时间复杂度： O(n²)
*/
const getChildren = root => {
root.children = arr.filter(v => v.pid === root.id)
root.children.forEach(v => getChildren(v))
}
const transferTree = arr => {
if (!Array.isArray(arr)) throw new Error('parameter arr must be Array.')
if (arr.length === 1) return arr
// 父子关系可以用map存储
const idMap = arr.reduce((total, cur) => Object.assign(total, { [cur.id]: { ...cur, children: [] } }), {})
// 找出所有根节点
const roots = arr.filter(node => !idMap[node.pid])
// 找出根节点下面所有子节点
roots.forEach(root => getChildren(root))
return roots
}
/*
方案2 非递归方式
时间复杂度： O(n)
*/
const transferTree2 = arr => {
if (!Array.isArray(arr)) throw new Error('parameter arr must be Array.')
if (arr.length === 1) return arr
const res = []
const idMap = arr.reduce((total, cur) => Object.assign(total, { [cur.id]: { ...cur, children: [] } }), {})
for (const [, node] of Object.entries(idMap)) {
// 根节点
if (idMap[node.pid]) {
res.push(node)
continue;
}
// 子节点
idMap[node.pid].children.push(node)
}
return res
}
console.log(transferTree2(arr))
