Javascript Run Recursively Array

Question

I'm trying to run recursive-ly in an array in a react aplication, I got it this way and I want to improve my function to run the array:

my Array:

tree: [
         {
           key: 0 ,
           name: 'menu 1',
           children: [
             {
               key: 1,
               name: 'submenu 1 - 1'
             },{
               key: 2,
               name: 'submenu 1 - 2'
             }
           ]
         },{
           key: 3 ,
           name: 'menu 2',
           children: [
             {
               key: 4,
               name: 'submenu 2 - 1'
             },{
               key: 5,
               name: 'submenu 2 - 2'
             }
           ]
         }
      ]

And I run this Array this way:

const findInArrayTree = (tree, search) =>{
  let it,
      result

  for (it = 0; it < tree.length; it += 1) {
    result = findNode(tree[it], search);
    if(result !== false){
      return result;
    }
  }
  return result;
};

const findNode = (currentNode, search) => {
    let i,
        currentChild,
        result; 
    if (search == currentNode.key) {
        return currentNode;
    } else {
        // Use a for loop instead of forEach to avoid nested functions
        // Otherwise "return" will not work properly
        if(currentNode.children){
          for (i = 0; i < currentNode.children.length; i += 1) {
            currentChild = currentNode.children[i];
            // Search in the current child
            result = findNode(currentChild, search);
            // Return the result if the node has been found
            if (result !== false) {
                return result;
            }
          }
        }
        // The node has not been found and we have no more options
        return false;
    }
};

Someone can help-me to improve this? I want to do this run in one function only, but my brain is broken alredy..

And what the value of findInArrayTree should be if, for instance, key = 3 ? — Faly
– Faly, Commented Oct 23, 2017 at 13:55
if key = 3 return: { key: 3 , name: 'menu 2', children: [ { key: 4, name: 'submenu 2 - 1' },{ key: 5, name: 'submenu 2 - 2' } ] } — bugalaws
– bugalaws, Commented Oct 23, 2017 at 14:15

Faly · Accepted Answer · 2017-10-23 14:08:25Z

1

You can do it like below:

var tree = [
         {
           key: 0 ,
           name: 'menu 1',
           children: [
             {
               key: 1,
               name: 'submenu 1 - 1'
             },{
               key: 2,
               name: 'submenu 1 - 2'
             }
           ]
         },{
           key: 3 ,
           name: 'menu 2',
           children: [
             {
               key: 4,
               name: 'submenu 2 - 1'
             },{
               key: 5,
               name: 'submenu 2 - 2'
             }
           ]
         }
      ];

function findInTree(items, search) {
    for (var i = 0; i < items.length; i++) {
        var item = items[i];
        if (item.key === search) {
            return item;
        } else if (item.children && item.children.length > 0) {
            var foundInChildren = findInTree(item.children, search);
            if (foundInChildren) {
                return foundInChildren;
            }
        }
    }
}

var found = findInTree(tree, 4);
console.log(found);

answered Oct 23, 2017 at 14:08

Faly

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1 Comment

bugalaws Over a year ago

Your sugestion works perfectly... thanks a lot friend!!

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