How to find a node in a tree with JavaScript

Question

I have and object literal that is essentially a tree that does not have a fixed number of levels. How can I go about searching the tree for a particualy node and then return that node when found in an effcient manner in javascript?

Essentially I have a tree like this and would like to find the node with the title 'randomNode_1'

var data = [
{
title: 'topNode',
 children: [
   {
       title: 'node1',
       children: [
       {
           title: 'randomNode_1'
       },
       {   
           title: 'node2',
           children: [
           {
               title: 'randomNode_2',
               children:[
               {   
                   title: 'node2',
                   children: [
                   {
                       title: 'randomNode_3',
                   }]
               }
               ]
           }]
       }]
   }
  ]
 }];

Answer 1

no BS version:

const find = (root, title) => 
  root.title === title ?
    root :
    root.children?.reduce((result, n) => result || find(n, title), undefined)

Usage:

const data = {
  title: "root",
  children: [
    { title: "ch1" },
    { title: "ch2", children: [{ title: "ch21" }] }
  ]
}

find(data, 'ch21') // returns { title: 'ch21' }

Answer 2

const flattenTree = (data: any) => {
  return _.reduce(
    data,
    (acc: any, item: any) => {
      acc.push(item);
      if (item.children) {
        acc = acc.concat(flattenTree(item.children));
        delete item.children;
      }
     return acc;
    },
    []
    );
   };

An Approach to convert the nested tree into an object with depth 0. We can convert the object in an object like this and can perform search more easily.

Answer 3

In 2022 use TypeScript and ES5

Just use basic recreation and built-in array method to loop over the array. Don't use Array.find() because this it will return the wrong node. Use Array.some() instead which allow you to break the loop.

 interface iTree {
     id: string;
     children?: iTree[];
 }

function findTreeNode(tree: iTree, id: string) {
    let result: iTree | null = null;

    if (tree.id === id) {
        result = tree;
    } else if (tree.children) {
        tree.children.some((node) => {
            result = findTreeNode(node, id);
            return result; // break loop
        });
    }

    return result;
}

Answer 4

FIND A NODE IN A TREE :

let say we have a tree like

let tree = [{
  id: 1,
  name: 'parent',
  children: [
    {
      id: 2,
      name: 'child_1'
    },
    {
      id: 3,
      name: 'child_2',
      children: [
        {
          id: '4',
          name: 'child_2_1',
          children: []
        },
        {
          id: '5',
          name: 'child_2_2',
          children: []
        }
      ]
    }
  ]
}];


function findNodeById(tree, id) {

   let result = null
   if (tree.id === id) {
        return tree;
   }

   if (Array.isArray(tree.children) && tree.children.length > 0) {
      tree.children.some((node) => {
        result = findNodeById(node, id);
        return result;
      });
   }
   return result;}

Answer 5

Wrote another one based on my needs

1. condition is injected.
2. path of found branch is available
3. current path could be used in condition statement
4. could be used to map the tree items to another object

// if predicate returns true, the search is stopped
function traverse2(tree, predicate, path = "") {
  if (predicate(tree, path)) return true;
  for (const branch of tree.children ?? [])
    if (traverse(branch, predicate, `${path ? path + "/" : ""}${branch.name}`))
      return true;
}

example

let tree = {
  name: "schools",
  children: [
    {
      name: "farzanegan",
      children: [
        {
          name: "classes",
          children: [
            { name: "level1", children: [{ name: "A" }, { name: "B" }] },
            { name: "level2", children: [{ name: "C" }, { name: "D" }] },
          ],
        },
      ],
    },
    { name: "dastgheib", children: [{ name: "E" }, { name: "F" }] },
  ],
};

traverse(tree, (branch, path) => {
  console.log("searching ", path);
  if (branch.name === "C") {
    console.log("found ", branch);
    return true;
  }
});

output

searching  
searching  farzanegan
searching  farzanegan/classes
searching  farzanegan/classes/level1
searching  farzanegan/classes/level1/A
searching  farzanegan/classes/level1/B
searching  farzanegan/classes/level2
searching  farzanegan/classes/level2/C
found  { name: 'C' }

Answer 6

Enhancement of the code based on "Erick Petrucelli"

1. Remove the 'reverse' option
2. Add multi-root support
3. Add an option to control the visibility of 'children'
4. Typescript ready
5. Unit test ready

function searchTree(
  tree: Record<string, any>[],
  value: unknown,
  key = 'value',
  withChildren = false,
) {
  let result = null;
  if (!Array.isArray(tree)) return result;

  for (let index = 0; index < tree.length; index += 1) {
    const stack = [tree[index]];
    while (stack.length) {
      const node = stack.shift()!;
      if (node[key] === value) {
        result = node;
        break;
      }
      if (node.children) {
        stack.push(...node.children);
      }
    }
    if (result) break;
  }
  if (withChildren !== true) {
    delete result?.children;
  }

  return result;
}

And the tests can be found at: https://gist.github.com/aspirantzhang/a369aba7f84f26d57818ddef7d108682

Answer 7

ES6+:

const deepSearch = (data, value, key = 'title', sub = 'children', tempObj = {}) => {
    if (value && data) {
        data.find((node) => {
            if (node[key] == value) {
                tempObj.found = node;
                return node;
            }
            return deepSearch(node[sub], value, key, sub, tempObj);
        });
        if (tempObj.found) {
            return tempObj.found;
        }
    }
    return false;
};

const result = deepSearch(data, 'randomNode_1', 'title', 'children');

Answer 8

Here's an iterative function using the Stack approach, inspired by FishBasketGordo's answer but taking advantage of some ES2015 syntax to shorten things.

Since this question has already been viewed a lot of times, I've decided to update my answer to also provide a function with arguments that makes it more flexible:

function search (tree, value, key = 'id', reverse = false) {
  const stack = [ tree[0] ]
  while (stack.length) {
    const node = stack[reverse ? 'pop' : 'shift']()
    if (node[key] === value) return node
    node.children && stack.push(...node.children)
  }
  return null
}

This way, it's now possible to pass the data tree itself, the desired value to search and also the property key which can have the desired value:

search(data, 'randomNode_2', 'title')

Finally, my original answer used Array.pop which lead to matching the last item in case of multiple matches. In fact, something that could be really confusing. Inspired by Superole comment, I've made it use Array.shift now, so the first in first out behavior is the default.

If you really want the old last in first out behavior, I've provided an additional arg reverse:

search(data, 'randomNode_2', 'title', true)

Answer 9

This is an iterative breadth first search. It returns the first node that contains a child of a given name (nodeName) and a given value (nodeValue).

getParentNode(nodeName, nodeValue, rootNode) {
    const queue= [ rootNode ]
    while (queue.length) {
        const node = queue.shift()
        if (node[nodeName] === nodeValue) {
            return node
        } else if (node instanceof Object) {
            const children = Object.values(node)
            if (children.length) {
                queue.push(...children)
            }
        }
    }
    return null
}

It would be used like this to solve the original question:

getParentNode('title', 'randomNode_1', data[0])

Answer 10

here is a more complex option - it finds the first item in a tree-like node with providing (node, nodeChildrenKey, key/value pairs & optional additional key/value pairs)

const findInTree = (node, childrenKey, key, value,  additionalKey?, additionalValue?) => {
  let found = null;
  if (additionalKey && additionalValue) {
    found = node[childrenKey].find(x => x[key] === value && x[additionalKey] === additionalValue);
  } else {
    found = node[childrenKey].find(x => x[key] === value);
  }
  if (typeof(found) === 'undefined') {
    for (const item of node[childrenKey]) {
      if (typeof(found) === 'undefined' && item[childrenKey] && item[childrenKey].length > 0) {
        found = findInTree(item, childrenKey, key, value, additionalKey, additionalValue);
      }
    }
  }
  return found;
};

export { findInTree };

Hope it helps someone.

Answer 11

find all parents of the element in the tree

let objects = [{
      id: 'A',
      name: 'ObjA',
      children: [
        {
          id: 'A1',
          name: 'ObjA1'
        },
        {
          id: 'A2',
          name: 'objA2',
          children: [
            {
              id: 'A2-1',
              name: 'objA2-1'
            },
            {
              id: 'A2-2',
              name: 'objA2-2'
            }
          ]
        }
      ]
    },
    {
      id: 'B',
      name: 'ObjB',
      children: [
        {
          id: 'B1',
          name: 'ObjB1'
        }
      ]
    }
    ];

let docs = [
  {

    object: {
      id: 'A',
      name: 'docA'
    },
    typedoc: {
      id: 'TD1',
      name: 'Typde Doc1'
    }
  },
  {

    object: {
      id: 'A',
      name: 'docA'
    },
    typedoc: {
      id: 'TD2',
      name: 'Typde Doc2'
    }
  },
  {

    object: {
      id: 'A1',
      name: 'docA1'
    },
    typedoc: {
      id: 'TDx1',
      name: 'Typde Doc x1'
    }
  },
  {

    object: {
      id: 'A1',
      name: 'docA1'
    },
    typedoc: {
      id: 'TDx2',
      name: 'Typde Doc x1'
    }
  },
  {

    object: {
      id: 'A2',
      name: 'docA2'
    },
    typedoc: {
      id: 'TDx2',
      name: 'Type de Doc x2'
    }
  },
  {

    object: {
      id: 'A2-1',
      name: 'docA2-1'
    },
    typedoc: {
      id: 'TDx2-1',
      name: 'Type de Docx2-1'
    },
  },
  {

    object: {
      id: 'A2-2',
      name: 'docA2-2'
    },
    typedoc: {
      id: 'TDx2-2',
      name: 'Type de Docx2-2'
    },
  },
  {

    object: {
      id: 'B',
      name: 'docB'
    },
    typedoc: {
      id: 'TD1',
      name: 'Typde Doc1'
    }
  },
  {

    object: {
      id: 'B1',
      name: 'docB1'
    },
    typedoc: {
      id: 'TDx1',
      name: 'Typde Doc x1'
    }

  }
];



function buildAllParents(doc, objects) {
    for (let o = 0; o < objects.length; o++) {
      let allParents = [];
      let getAllParents = (o, eleFinded) => {
        if (o.id === doc.object.id) {
          doc.allParents = allParents;
          eleFinded = true;
          return { doc, eleFinded };
        }
        if (o.children) {
          allParents.push(o.id);
          for (let c = 0; c < o.children.length; c++) {
            let { eleFinded, doc } = getAllParents(o.children[c], eleFinded);
            if (eleFinded) {
              return { eleFinded, doc };
            } else {
              continue;
            }
          }

        }
        return { eleFinded };
      };
      if (objects[o].id === doc.object.id) {
        doc.allParents = [objects[o].id];
        return doc;
      } else if (objects[o].children) {
        allParents.push(objects[o].id);
        for (let c = 0; c < objects[o].children.length; c++) {
          let eleFinded = null;`enter code here`
          let res = getAllParents(objects[o].children[c], eleFinded);
          if (res.eleFinded) {
            return res.doc;
          } else {
            continue;
          }
        }
      }

    }
  }

docs = docs.map(d => buildAllParents(d, objects`enter code here`))

Answer 12

Here's an iterative solution:

var stack = [], node, ii;
stack.push(root);

while (stack.length > 0) {
    node = stack.pop();
    if (node.title == 'randomNode_1') {
        // Found it!
        return node;
    } else if (node.children && node.children.length) {
        for (ii = 0; ii < node.children.length; ii += 1) {
            stack.push(node.children[ii]);
        }
    }
}

// Didn't find it. Return null.
return null;

Answer 13

A flexible recursive solution that will work for any tree

// predicate: (item) => boolean
// getChildren: (item) => treeNode[]
searchTree(predicate, getChildren, treeNode) {
        function search(treeNode) {
            if (!treeNode) {
                return undefined;
            }

            for (let treeItem of treeNode) {
                if (predicate(treeItem)) {
                    return treeItem;
                }

                const foundItem = search(getChildren(treeItem));

                if (foundItem) {
                    return foundItem;
                }
            }
        }

        return search(treeNode);
    }

Answer 14

This function is universal and does search recursively. It does not matter, if input tree is object(single root), or array of objects (many root objects). You can configure prop name that holds children array in tree objects.

// Searches items tree for object with specified prop with value
// 
// @param {object} tree nodes tree with children items in nodesProp[] table, with one (object) or many (array of objects) roots
// @param {string} propNodes name of prop that holds child nodes array
// @param {string} prop name of searched node's prop
// @param {mixed} value value of searched node's  prop
// @returns {object/null} returns first object that match supplied arguments (prop: value) or null if no matching object was found

function searchTree(tree, nodesProp, prop, value) {
  var i, f = null; // iterator, found node
  if (Array.isArray(tree)) { // if entry object is array objects, check each object
    for (i = 0; i < tree.length; i++) {
      f = searchTree(tree[i], nodesProp, prop, value);
      if (f) { // if found matching object, return it.
        return f;
      }
    }
  } else if (typeof tree === 'object') { // standard tree node (one root)
    if (tree[prop] !== undefined && tree[prop] === value) {
      return tree; // found matching node
    }
  }
  if (tree[nodesProp] !== undefined && tree[nodesProp].length > 0) { // if this is not maching node, search nodes, children (if prop exist and it is not empty)
    return searchTree(tree[nodesProp], nodesProp, prop, value);
  } else {
    return null; // node does not match and it neither have children
  }
}

I tested it localy and it works ok, but it somehow won't run on jsfiddle or jsbin...(recurency issues on those sites ??)

run code :

    var data = [{
      title: 'topNode',
      children: [{
        title: 'node1',
        children: [{
          title: 'randomNode_1'
        }, {
          title: 'node2',
          children: [{
            title: 'randomNode_2',
            children: [{
              title: 'node2',
              children: [{
                title: 'randomNode_3',
              }]
            }]
          }]
        }]
      }]
    }];

    var r = searchTree(data, 'children', 'title', 'randomNode_1');
    //var r = searchTree(data, 'children', 'title', 'node2');  // check it too
    console.log(r);

It works in http://www.pythontutor.com/live.html#mode=edit (paste the code)

Answer 15

Basing this answer off of @Ravindra's answer, but with true recursion.

function searchTree(element, matchingTitle){
     if(element.title == matchingTitle){
          return element;
     }else if (element.children != null){
          var i;
          var result = null;
          for(i=0; result == null && i < element.children.length; i++){
               result = searchTree(element.children[i], matchingTitle);
          }
          return result;
     }
     return null;
}

Then you could call it:

var element = data[0];
var result = searchTree(element, 'randomNode_1');

Answer 16

My answer is inspired from FishBasketGordo's iterativ answer. It's a little bit more complex but also much more flexible and you can have more than just one root node.

/**searchs through all arrays of the tree if the for a value from a property
 * @param aTree : the tree array
 * @param fCompair : This function will receive each node. It's upon you to define which 
                     condition is necessary for the match. It must return true if the condition is matched. Example:
                        function(oNode){ if(oNode["Name"] === "AA") return true; }
 * @param bGreedy? : us true to do not stop after the first match, default is false
 * @return an array with references to the nodes for which fCompair was true; In case no node was found an empty array
 *         will be returned
*/
var _searchTree = function(aTree, fCompair, bGreedy){
    var aInnerTree = []; // will contain the inner children
    var oNode; // always the current node
    var aReturnNodes = []; // the nodes array which will returned

    // 1. loop through all root nodes so we don't touch the tree structure
    for(keysTree in aTree) {
        aInnerTree.push(aTree[keysTree]);
    }
    while(aInnerTree.length > 0) {
        oNode = aInnerTree.pop();
        // check current node
        if( fCompair(oNode) ){
            aReturnNodes.push(oNode);
            if(!bGreedy){
                return aReturnNodes;
            }
        } else { // if (node.children && node.children.length) {
            // find other objects, 1. check all properties of the node if they are arrays
            for(keysNode in oNode){
                // true if the property is an array
                if(oNode[keysNode] instanceof Array){
                    // 2. push all array object to aInnerTree to search in those later
                    for (var i = 0; i < oNode[keysNode].length; i++) {
                        aInnerTree.push(oNode[keysNode][i]);
                    }
                }
            }
        }
    }
    return aReturnNodes; // someone was greedy
}

Finally you can use the function like this:

var foundNodes = _searchTree(data, function(oNode){ if(oNode["title"] === "randomNode_3") return true; }, false);
console.log("Node with title found: ");
console.log(foundNodes[0]);

And if you want to find all nodes with this title you can simply switch the bGreedy parameter:

var foundNodes = _searchTree(data, function(oNode){ if(oNode["title"] === "randomNode_3") return true; }, true);
console.log("NodeS with title found: ");
console.log(foundNodes);

Answer 17

This is basic recursion problem.

window.parser = function(searchParam, data) {
  if(data.title != searchParam) {
    returnData = window.parser(searchParam, children)
  } else {
     returnData = data;
  }

  return returnData;
}

Answer 18

The following is working at my end:

function searchTree(data, value) {
if(data.title == value) {
    return data;
}
if(data.children && data.children.length > 0) {
    for(var i=0; i < data.children.length; i++) {
        var node = traverseChildren(data.children[i], value);
        if(node != null) {
            return node;
        }
    }
}
return null;

}

Answer 19

You have to use recursion.

var currChild = data[0];
function searchTree(currChild, searchString){
     if(currChild.title == searchString){
          return currChild;
     }else if (currChild.children != null){
          for(i=0; i < currChild.children.length; i ++){
               if (currChild.children[i].title ==searchString){
                    return currChild.children[i];
               }else{
                    searchTree(currChild.children[i], searchString);
               }
          }
          return null;
     }
     return null;
}