how to find object using recursion in javascript?

Question

let tree = {
    name: "A",
    children: [
        {
            name: 'A-1',
            children: [
                {name: "A-1-A"},
                {name: "A-1-B"},
            ]
        },
        {
            name: 'B-1',
            children: [
                {
                    name: "B-1-A",
                    children: [
                        {name: "B-11-A"},
                        {name: "B-11-B"}
                    ]
                },
                {name: "B-1-B"},
            ]
        },

    ]


};

I am trying to find object from tree object using recursion .

when I call like this searchFn(tree,'A-1') it should return { name: 'A-1', children: [ {name: "A-1-A"}, {name: "A-1-B"}, ] } object

it I call like this searchFn(tree,'A-1-A') it should return this

{name: "A-1-A"}

I tried like this but not working

 function searchFn(obj ,searchText){
        if(obj.name === searchText) return obj
        if(obj.children.length > 0){
          return   searchFn(obj.children.pop(),searchText)
        }
      return null
    }

Nina Scholz · Accepted Answer · 2020-04-02 12:38:44Z

4

You need to iterate the children of the object and take a variable for the result.

function searchFn(object, searchText) {
    var result;
    if (object.name === searchText) return object;
    (object.children || []).some(o => result = searchFn(o, searchText));
    return result || null;
}


let tree = { name: "A", children: [{ name: 'A-1', children: [{ name: "A-1-A" }, { name: "A-1-B" }] }, { name: 'B-1', children: [{ name: "B-1-A", children: [{ name: "B-11-A" }, { name: "B-11-B" }] }, { name: "B-1-B" }] }] };

console.log(searchFn(tree, 'foo'));
console.log(searchFn(tree, 'A-1'));

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answered Apr 2, 2020 at 12:38

Nina Scholz

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Comments

Alexandre Senges · Accepted Answer · 2020-04-02 13:00:52Z

2

First, you should create a class for readability. Then you should have a loop inside the function that iterates over the children and only returns if a child is found.

    class Node {
        static fromStruct = (struct) => {
            const tree = new Node(struct.name);
            if(!struct.children) return tree;
            for(const child of struct.children) {
                tree.children.push(Node.fromStruct(child));
            }
            return tree;
        }

        constructor(name){
            this.name = name;
            this.children = [];
        }
    
        addChild = (parentName, childName) => {
            const parent = this.searchFn(parentName);
            if(!parent) throw new Error("Parent not found");
            parent.children.push(new Node(childName));
        }
    
        searchFn = (name) => {
            if(this.name === name) return this;
            for(const child of this.children) {
                const found = child.searchFn(name);
                if(found !== null) return found;
            }
            return null;
        }
    }


    const data = {
        name: "A",
        children: [
            {
                name: 'A-1',
                children: [
                    {name: "A-1-A"},
                    {name: "A-1-B"},
                ]
            },
            {
                name: 'B-1',
                children: [
                    {
                        name: "B-1-A",
                        children: [
                            {name: "B-11-A"},
                            {name: "B-11-B"}
                        ]
                    },
                    {name: "B-1-B"},
                ]
            },
    
        ]
    };

    const tree = Node.fromStruct(data);
    console.log(tree.searchFn("A-1-A"));

edited Apr 2, 2020 at 13:00

answered Apr 2, 2020 at 12:38

Alexandre Senges

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2 Comments

user944513 Over a year ago

can you please run with sample

Alexandre Senges Over a year ago

@user944513 I improved it a lot

xdeepakv · Accepted Answer · 2020-04-02 12:47:29Z

1

const tree = {"name":"A","children":[{"name":"A-1","children":[{"name":"A-1-A"},{"name":"A-1-B"}]},{"name":"B-1","children":[{"name":"B-1-A","children":[{"name":"B-11-A"},{"name":"B-11-B"}]},{"name":"B-1-B"}]}]}

const searchFn = (tree, name) => {
  let result = null;
  if (typeof tree !== "object") return result;
  if (tree.name === name) return tree;
  if (tree.children && tree.children.length) {
    tree.children.some(data => (result = searchFn(data, name)));
  }
  return result;
};
console.log(searchFn(tree, "A-1-A"));
console.log(searchFn(tree, "A-1"));
console.log(searchFn(tree, ""));

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answered Apr 2, 2020 at 12:47

xdeepakv

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Comments

Code Maniac · Accepted Answer · 2020-04-02 12:52:10Z

1

You can try something like this

let tree = { name: "A", children: [{ name: 'A-1', children: [{ name: "A-1-A" }, { name: "A-1-B" }] }, { name: 'B-1', children: [{ name: "B-1-A", children: [{ name: "B-11-A" }, { name: "B-11-B" }] }, { name: "B-1-B" }] }] };

function searchFn(obj, text){
  if(obj.name === text) return obj
  else{
    if(obj && obj.children && Array.isArray(obj.children)){
      for(let value of obj.children){
        if(searchFn(value, text)){
          return searchFn(value, text)
        }
      }
    }
  }
  return null
}

console.log(searchFn(tree, 'A-1'))
console.log(searchFn(tree, 'A-1-A'))
console.log(searchFn(tree, 'A-A-A'))

answered Apr 2, 2020 at 12:52

Code Maniac

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Comments

Param Siddharth · Accepted Answer · 2020-04-02 12:55:24Z

0

The problem with your function is that it only checks for the popped element from the array in the children property of the passed object, and not the other elements. Try this:

function searchFn(obj ,searchText) { 
    if(obj.name === searchText) return obj; 
    if(obj.children) {
        for (let x of obj.children) {
            let y = searchFn(x,searchText);
            if (y)
                return y;
        }
    }
    return null; 
}

answered Apr 2, 2020 at 12:55

Param Siddharth

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