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Added parentNode and minor cleanup

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This commit is contained in:
Lakshya Thakur
2020-11-09 00:05:57 +05:30
parent 05fa204900
commit dea8bcafd8
2 changed files with 39 additions and 10 deletions
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docs/element-child-content.mdx
+37 -8
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@@ -34,20 +34,49 @@ Text nodes are not elements, so they don't have any tagName. To identify the typ
<MDN title="nodeType" i url="https://developer.mozilla.org/en-US/docs/Web/API/Node/nodeType" />&nbsp; <MDN title="nodeType" i url="https://developer.mozilla.org/en-US/docs/Web/API/Node/nodeType" />&nbsp;
</span> </span>
### 2. parentElement ### 2. parentNode
`Node.parentElement: Element | null` `Node.parentNode: Element | Document | DocumentFragment | null`
This property returns **the parent element** of the current node. If there is no parent, it returns null. This property returns **the parent node** of the current node. If there is no parent, it returns null (which is always true in the case of Document and DocumentFragment).
The topmost parent element in DOM is `document`. You can use this to check if an element is detached from dom.
```js ```js
<div id="parent"> <div id="parent">
<div id="child"/> <div id="child"/>
</div> </div>
child.parentElement // <div id="parent>...</div>" child.parentNode // <div id="parent>...</div>
``` ```
<Accordion title="Related">
<Admonition type="note" title="Node.parentElement : Element | null">
This property returns ** the parent element ** of the current node. If there is no parent, it returns null.
The topmost parent node in DOM is document. This can be used to check if an element is detached from DOM.
```js
let newEle = document.createElement("div");
newEle.parentNode // null
newEle.parentElement // null
// In case of appending to an element
document.body.append(newEle)
newEle.parentNode // <body>...</body>
newEle.parentElement // <body>...</body>
//In case of appending to a document fragment, say, df
df.append(newEle)
newEle.parentNode // document-fragment
newEle.parentElement // null
```
</Admonition>
</Accordion>
<span><MDN title="parentNode" url="https://developer.mozilla.org/en-US/docs/Web/API/Node/parentNode" /></span>
## Children ## Children
@@ -194,19 +223,19 @@ There is also a similar `previousSibling` property which returns the previous no
This allows accessing the DOM tree in [Left-child right-sibling](https://en.wikipedia.org/wiki/Left-child_right-sibling_binary_tree) (LCRS) form, which is more memory efficient for usual traversal. This allows accessing the DOM tree in [Left-child right-sibling](https://en.wikipedia.org/wiki/Left-child_right-sibling_binary_tree) (LCRS) form, which is more memory efficient for usual traversal.
#### How do you implement **dom traversal**? #### How do you implement dom traversal ?
The typical approach is to use a queue, push all children to the queue, pop them one-by-one and visit them. This requires **extra memory** because of the queue data structure which holds all the *next items to visit*. The typical approach is to use a queue, push all children to the queue, pop them one-by-one and visit them. This requires **extra memory** because of the queue data structure which holds all the *next items to visit*.
But with the LCRS form, you always have access to the next item **without using extra space**. If current element has child, visit `.firstChild`. Then visit next node using `.nextSibling`. But with the LCRS form, you always have access to the next item **without using extra space**. If current element has child, visit `.firstChild`. Then visit next node using `.nextSibling`.
```js ```js
// 1. Typical approach // 1. Typical approach
var queue = [root] let queue = [root]
while (queue.length > 0) { while (queue.length > 0) {
const item = queue.shift(); const item = queue.shift();
console.log(item); console.log(item);
if (item.hasChildNodes()) { if (item.hasChildNodes()) {
queue = Array.from(item.childNodes).concat(queue) queue.push(Array.from(item.childNodes));
} }
} }
docusaurus.config.js
+1 -1
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@@ -1,6 +1,6 @@
module.exports = { module.exports = {
title: 'UI Questions', title: 'UI Questions',
tagline: 'Review all frontend topics before a interview', tagline: 'Review all frontend topics before an interview',
url: 'https://UiQuestions.com', url: 'https://UiQuestions.com',
baseUrl: '/', baseUrl: '/',
onBrokenLinks: 'throw', onBrokenLinks: 'throw',