How to convert a tree into an array? - arrays

Question

I have a tree object which is an irregular tree which children's names and key values can change everytime I run my code. For example:
{
addressRouter: 192.168.0.1,
addresses:
{
address1: 'A',
},
{
address2: 'B',
},
{
ports: [
{
portA: 'C',
portB: null
},
}
route: 'D',
}
so the names: 'addressRouter', 'addresses', 'address1', etc and their keys are unpredictable but I need to convert the tree object in arrays with the following format:
addressRouter
addresses/address1
addresses/address2
addresses/ports/portA
addresses/ports/portB
route
and then have their keys next to them.
I have this function to construct the tree, which is correct:
const iterate = (obj, obj2) => {
Object.keys(obj).forEach(key => {
obj2[key] = obj[key];
if (typeof obj[key] === 'object') {
iterate(obj[key], obj2)
}
})
}
but after debugging, I realized it doesn't get all branches.

Answer 1

We can use a recursive function to traverse the tree and get the keys in the required format.
I am assuming addresses in the given tree object is an array of objects
function processTree(obj, rootKey) {
const arr = [];
obj && Object.keys(obj).forEach(key => {
const val = obj[key];
if (val && val instanceof Array) {
val.forEach(item => arr.push(...processTree(item, key)))
}else if (val && typeof(val) == "object") {
arr.push(...processTree(val, key));
}else {
arr.push(key);
}
});
return rootKey ? arr.map(item => rootKey + "/" + item) : arr;
}
console.log(processTree(tree, null));
Result : ["addressRouter", "addresses/address1", "addresses/address2", "addresses/ports/portA", "addresses/ports/portB", "route"]

Answer 2

I just write the code below, see if it's what you want.
const tree = {
addressRouter: '192.168.0.1',
addresses: [
{
address1: 'A',
},
{
address2: 'B',
},
{
ports: [
{
portA: 'C',
portB: null,
},
],
},
],
route: 'D',
};
const traverse = (input) => {
const resultList = [];
const isEndPoint = (obj) => {
return typeof obj !== 'object' || obj === null;
};
const buildPath = (currentPath, key) =>
currentPath === '' ? key : `${currentPath}/${key}`;
const innerTraverse = (tree, currentPath = '') => {
if (tree !== null && typeof tree === 'object') {
Object.entries(tree).forEach(([key, value]) => {
if (isEndPoint(value)) {
resultList.push(buildPath(currentPath, key));
return;
}
let path = currentPath;
if (!Array.isArray(tree)) {
path = buildPath(currentPath, key);
}
innerTraverse(value, path);
});
}
};
innerTraverse(input);
return resultList;
};
console.log(traverse(tree));
/**
* [
* 'addressRouter',
* 'addresses/address1',
* 'addresses/address2',
* 'addresses/ports/portA',
* 'addresses/ports/portB',
* 'route'
* ]
*/

Answer 3

I've just found the loop I needed it here, which is:
function* traverse(o,path=[]) {
for (var i of Object.keys(o)) {
const itemPath = path.concat(i);
yield [i,o[i],itemPath];
if (o[i] !== null && typeof(o[i])=="object") {
//going one step down in the object tree!!
yield* traverse(o[i],itemPath);
}
}
}
Then, if the tree object (first gray box in my question) were name, for instance, "params", I would do this:
if (params != null && params != undefined) {
for(var [key, value, path] of traverse(params)) {
// do something here with each key and value
if (typeof value == 'string'){
var tempName = '';
for (name in path) {
//console.log(path[name])
p= tempName += path[name] + "/"
}
console.log(p, value)
}
}
}