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Lakshya Thakur
Lakshya Thakur

Posted on • Originally published at blog.lakbychance.com

Implementing the DOM contains() method

As per MDN,

The Node.contains() method returns a Boolean value indicating whether a node is a descendant of a given node, i.e. the node itself, one of its direct children (childNodes), one of the children's direct children, and so on.

But wait, Node.prototype.contains(...) already exists. I want another name for our custom function. Let's google synonym of contains coz

why not

Intense googling later......

Screen Shot 2021-07-06 at 7.26.12 PM.png

Certainly we ain't going with swallow. I think includes would be cool as both Array and String have includes as well in their prototypes.

inclusion

Before we proceed one important thing to know is that when adding new method to prototype and expecting to use it like so :-

document.includes(document.body),

the method should not be an arrow function so that document can be accessed inside the includes function via this keyword.

Alright then, let's implement Node.prototype.includes in 4 different ways :-

four

The recursive DFS

1 Node.prototype.includes = function(node){
2 const currentNode = this;
3  if(!currentNode)
4   return false;
5  if(currentNode===node)
6   return true;
7  let isNodeFound = false;
8 for(let index = 0;index<currentNode.childNodes.length;index++){
9    isNodeFound = isNodeFound || currentNode.childNodes[index].includes(node);
10   if(isNodeFound) return true;
11  }
12  return false;
13 }
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Explanation :-

  • Line 2 to 4 - Set currentNode to this and If currentNode doesn't exist, simply return false.
  • Line 5 to 6 - if currentNode is equal to node return true.
  • Line 7 to 13 - Initialize isNodeFound to false. Then loop over childNodes of the currentNode and on each child, call the includes method again to check if they include the node element. If they do, isNodeFound will ultimately become true since it is being Orrrrrrd with the results coming from respective childNodes and reassigned to itself. Once isNodeFound is true, we don't need to loop over rest of the childNodes of currentNode and exit early by returning true else ultimately return false.

The iterative BFS

1 Node.prototype.includes = function (node) {
2 const queue = [];
3  let currentNode = this;
4  queue.push(currentNode);
5  while (queue.length) {
6    currentNode = queue.shift();
7    if (currentNode === node) return true;
8    if (currentNode.hasChildNodes()) {
9      queue.push(...currentNode.childNodes);
10    }
11 }
12  return false;
13 };
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Explanation :-

  • Line 2 to 4 - Initialize an empty list as queue. Set currentNode to this and push (or enqueue to be specific) it.
  • Line 5 to 12 - While the queue is not empty, dequeue the currentNode from front of the queue (using shift here). If currentNode is equal to node then return true. Otherwise enqueue the childNodes of currentNode (using push here). Once we are out of the while loop, we have traversed all the nodes and can safely say that we couldn't find the node and return false.

Note - The above can be transformed to iterative DFS by using pop instead of shift and obviously for the sake of consistency, rename queue to stack.

nerd

Till now both the approaches followed the classic DS/Algo traversal with DFS and BFS.
We are now going to see 2 more approaches which take benefit of certain properties that are specifically applicable to DOM nodes.


LCRS (Left Child Right Sibling) form

1 Node.prototype.includes = function (node) {
2 const currentNode = this;
3 if (!currentNode)
4   return false;
5 if (currentNode === node) return true;
6 return !!(currentNode.firstChild?.includes(node) || currentNode.nextSibling?.includes(node))
7 };
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Explanation :-

  • Line 2 to 5 -
    • Initialize currentNode to this and if currentNode doesn't exist, return false.
    • If currentNode is equal to node return true
  • Line 6 - DOM nodes not only have pointers to their childNodes but also to their sibling nodes as well as parent nodes. Here we are going to leverage the sibling factor for easy traversal. So, we can now check whether the current node's firstChild includes the node OR current node's nextSibling includes the node. Also notice the !!. That's because I have used the ? operator due to which we can end up with undefined || undefined condition or false || undefined condition where both evaluate to undefined which is a falsy value and so !! will ensure undefined coerces to false.

sibling


Using parentNode

1 Node.prototype.includes = function(node){
2 const currentNode = this;
3  while(node){
4    if(currentNode===node) return true;
5    node = node.parentNode;
6  }
7  return false;
8 }
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Explanation :-

  • Line 2 to 7 - Remember DOM node being so attached to it's siblings and parent ? The latter one works well for this use-case as well. While node exists, we check if currentNode is equal to node and if it is we return true, else the node is made to point to it's parentNode for further comparisons. If we exit the while loop, it's safe to say that the node isn't a contained within the currentNode and thus, return false.

parent

And here is a working codepen with all 4 implementations. Comment the rest for any one to reflect ✨.

Have more ways to implement the same ? Feel free to share your approach in the comment section 👇.

Thank you for your time :D

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