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:leaves: Agnostic tree traversal library.

License: MIT License

JavaScript 100.00%

tree-crawl's Introduction

tree-crawl travis codecov greenkeeper size

Agnostic tree traversal library.

  • Agnostic: Supports any kind of tree. You provide a way to access a node's children, that's it.
  • Fast: Crafted to be optimizer-friendly. See performance for more details.
  • Mutation friendly: Does not ๐Ÿ’ฅ when you mutate the tree.
  • Multiple orders: Supports DFS pre and post order and BFS traversals.

Quickstart

Installation

You can install tree-crawl with yarn:

$ yarn add tree-crawl

Alternatively using npm:

$ npm install --save tree-crawl

Usage

import crawl from 'tree-crawl'

// traverse the tree in pre-order
crawl(tree, console.log)
crawl(tree, console.log, { order: 'pre' })

// traverse the tree in post-order
crawl(tree, console.log, { order: 'post' })

// traverse the tree using `childNodes` as the children key
crawl(tree, console.log, { getChildren: node => node.childNodes })

// skip a node and its children
crawl(tree, (node, context) => {
  if ('foo' === node.type) {
    context.skip()
  }
})

// stop the walk
crawl(tree, (node, context) => {
  if ('foo' === node.type) {
    context.break()
  }
})

// remove a node
crawl(tree, (node, context) => {
  if ('foo' === node.type) {
    context.parent.children.splice(context.index, 1)
    context.remove()
  }
})

// replace a node
crawl(tree, (node, context) => {
  if ('foo' === node.type) {
    const node = {
      type: 'new node',
      children: [
        { type: 'new leaf' }
      ]
    }
    context.parent.children[context.index] = node
    context.replace(node)
  }
})

FAQ

How can I get the path of the current node (#37)?

tl;dr It's easy for DFS, less easy for BFS

If you are using DFS you can use the following utility function:

const = getPath(context) =>
  context.cursor.stack.xs.reduce((path, item) => {
    if (item.node) {
      path.push(item.node)
    }
    return path
  })

If you are really concerned about performance, you could read items from the stack directly. Each item has a node and index property that you can use. The first item in the stack can be discarded and will have a node set to null. Be aware that you should not mutate the stack, or it will break the traversal.

If you are using BFS, things gets more complex. A simple hacky way to do so is to traverse the tree using DFS first. You can ad a path property to your nodes using the method above. And then do your regular BFS traversal using that path property.

API

Iteratee

Called on each node of the tree.

Type: Function

Parameters

  • node Object Node being visited.
  • context Context Traversal context

Options

Walk options.

Type: Object

Parameters

  • node

Properties

  • getChildren Function? Return a node's children.
  • order ("pre" | "post" | "bfs")? Order of the walk either in DFS pre or post order, or BFS.

Examples

Traverse a DOM tree.

crawl(document.body, doSomeStuff, { getChildren: node => node.childNodes })

BFS traversal

crawl(root, doSomeStuff, { order: 'bfs' })

crawl

Walk a tree recursively.

By default getChildren will return the children property of a node.

Parameters

  • root Object Root node of the tree to be walked.
  • iteratee Iteratee Function invoked on each node.
  • options Options? Options customizing the walk.

Context

A traversal context.

Four operations are available. Note that depending on the traversal order, some operations have no effects.

Parameters

  • flags Flags
  • cursor Cursor

skip

Skip current node, children won't be visited.

Examples

crawl(root, (node, context) => {
  if ('foo' === node.type) {
    context.skip()
  }
})

break

Stop traversal now.

Examples

crawl(root, (node, context) => {
  if ('foo' === node.type) {
    context.break()
  }
})

remove

Notifies that the current node has been removed, children won't be visited.

Because tree-crawl has no idea about the intrinsic structure of your tree, you have to remove the node yourself. Context#remove only notifies the traversal code that the structure of the tree has changed.

Examples

crawl(root, (node, context) => {
  if ('foo' === node.type) {
    context.parent.children.splice(context.index, 1)
    context.remove()
  }
})

replace

Notifies that the current node has been replaced, the new node's children will be visited instead.

Because tree-crawl has no idea about the intrinsic structure of your tree, you have to replace the node yourself. Context#replace notifies the traversal code that the structure of the tree has changed.

Parameters

  • node Object Replacement node.

Examples

crawl(root, (node, context) => {
  if ('foo' === node.type) {
    const node = {
      type: 'new node',
      children: [
        { type: 'new leaf' }
      ]
    }
    context.parent.children[context.index] = node
    context.replace(node)
  }
})

parent

Get the parent of the current node.

Returns Object Parent node.

depth

Get the depth of the current node. The depth is the number of ancestors the current node has.

Returns Number Depth.

level

Get the level of current node. The level is the number of ancestors+1 the current node has.

Returns Number Level.

index

Get the index of the current node.

Returns Number Node's index.

Performance

tree-crawl is built to be super fast and traverse potentially huge trees. It's possible because it implements its own stack and queue for traversal algorithms and makes sure the code is optimizable by the VM.

If you do need real good performance please consider reading this checklist first.

Your main objective is to keep the traversal code optimized and avoid de-optimizations and bailouts. To do so, your nodes should have the same hidden class and your code stay monomorphic.

Related

License

MIT ยฉ Nicolas Gryman

tree-crawl's People

Contributors

ngryman avatar greenkeeper[bot] avatar greenkeeperio-bot avatar jameswlane avatar deltaidea avatar

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