我有一个来自NPM包'directory-tree'的目录结构,我希望将其压缩成一个更简单的嵌套结构。我想要一个尾递归解决方案将第一个对象转换为第二个对象,但是我无法绕过如何构造它。
当然,主要条件是第一个结构中的“节点”是“文件”还是“目录”。如果它是一个文件,我们只需要文件的基本名称来键入相对路径。但是,如果它是一个目录,我们希望目录的基本名称键入一个对象并在那里递归。
我将用他们的例子来说明结构:
{
"path": "photos",
"name": "photos",
"size": 600,
"type": "directory",
"children": [
{
"path": "photos/summer",
"name": "summer",
"size": 400,
"type": "directory",
"children": [
{
"path": "photos/summer/june",
"name": "june",
"size": 400,
"type": "directory",
"children": [
{
"path": "photos/summer/june/windsurf.jpg",
"name": "windsurf.jpg",
"size": 400,
"type": "file",
"extension": ".jpg"
}
]
}
]
},
{
"path": "photos/winter",
"name": "winter",
"size": 200,
"type": "directory",
"children": [
{
"path": "photos/winter/january",
"name": "january",
"size": 200,
"type": "directory",
"children": [
{
"path": "photos/winter/january/ski.png",
"name": "ski.png",
"size": 100,
"type": "file",
"extension": ".png"
},
{
"path": "photos/winter/january/snowboard.jpg",
"name": "snowboard.jpg",
"size": 100,
"type": "file",
"extension": ".jpg"
}
]
}
]
}
]
}
我希望最终结构更加简单。如下所示:
{
"photos": {
"summer": {
"june": {
"windsurf.jpg": "photos/summer/june/windsurf.jpg"
}
},
"winter": {
"january": {
"ski.png": "photos/winter/january/ski.png",
"snowboard.jpg": "photos/winter/january/snowboard.jpg"
}
}
}
}
答案 0 :(得分:1)
function copyNode(node, result = {}){
if(node.type === "directory"){
const folder = result[node.name] = {};
for(const sub of node.children)
copyNode(sub, folder);
} else {
result[node.name] = node.path;
}
return result;
}
这是一种简单的递归方法,这不是尾调用递归,因为仅使用一个尾调用遍历树是非常困难的(=不值得)。
答案 1 :(得分:1)
我们可以为您的案例将深度优先搜索转换为尾递归。
let testObj = {
"path": "photos",
"name": "photos",
"size": 600,
"type": "directory",
"children": [
{
"path": "photos/summer",
"name": "summer",
"size": 400,
"type": "directory",
"children": [
{
"path": "photos/summer/june",
"name": "june",
"size": 400,
"type": "directory",
"children": [
{
"path": "photos/summer/june/windsurf.jpg",
"name": "windsurf.jpg",
"size": 400,
"type": "file",
"extension": ".jpg"
}
]
}
]
},
{
"path": "photos/winter",
"name": "winter",
"size": 200,
"type": "directory",
"children": [
{
"path": "photos/winter/january",
"name": "january",
"size": 200,
"type": "directory",
"children": [
{
"path": "photos/winter/january/ski.png",
"name": "ski.png",
"size": 100,
"type": "file",
"extension": ".png"
},
{
"path": "photos/winter/january/snowboard.jpg",
"name": "snowboard.jpg",
"size": 100,
"type": "file",
"extension": ".jpg"
}
]
}
]
}
]
};
function tailRecurse(stack, result){
if (!stack.length)
return result;
// stack will contain
// the next object to examine
[obj, ref] = stack.pop();
if (obj.type == 'file'){
ref[obj.name] = obj.path;
} else if (obj.type == 'directory'){
ref[obj.name] = {};
for (let child of obj.children)
stack.push([child, ref[obj.name]]);
}
return tailRecurse(stack, result);
}
// Initialise
let _result = {};
let _stack = [[testObj, _result]];
console.log(tailRecurse(_stack, _result));

答案 2 :(得分:0)
你可以通过取消类型来采取递归方法。
对于'directory'
,取一个对象并迭代孩子。
否则,请指定具有给定名称的密钥的路径。
function fn(source, target) {
if (source.type === 'directory') {
target[source.name] = {};
(source.children || []).forEach(o => fn(o, target[source.name]));
} else {
target[source.name] = source.path;
}
}
var source = { path: "photos", name: "photos", size: 600, type: "directory", children: [{ path: "photos/summer", name: "summer", size: 400, type: "directory", children: [{ path: "photos/summer/june", name: "june", size: 400, type: "directory", children: [{ path: "photos/summer/june/windsurf.jpg", name: "windsurf.jpg", size: 400, type: "file", extension: ".jpg" }] }] }, { path: "photos/winter", name: "winter", size: 200, type: "directory", children: [{ path: "photos/winter/january", name: "january", size: 200, type: "directory", children: [{ path: "photos/winter/january/ski.png", name: "ski.png", size: 100, type: "file", extension: ".png" }, { path: "photos/winter/january/snowboard.jpg", name: "snowboard.jpg", size: 100, type: "file", extension: ".jpg" }] }] }] },
target = {};
fn(source, target);
console.log(target);

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