A binary tree is a tree where each node may only have up to two children. These children are stored on the leftand right properties of each node.

When traversing a binary tree, we have three common traversal algorithms: in order, pre-order, and post-order. In this lesson, we write each of these algorithms and explore their differences.

// Binary Trees and Tree Traversal

// Binary trees are trees whose nodes can only have up to two children

function createBinaryNode(key) {
return {
key,
left: null,
right: null,
addLeft(leftKey) {
const newLeft = createBinaryNode(leftKey)
this.left = newLeft
return newLeft
},
addRight(rightKey) {
const newRight = createBinaryNode(rightKey)
this.right = newRight
return newRight
}
}
} const TRAVERSALS = {
/**
* Javascript Call stack is Last in, First Out,
* So it keep calling
* TRAVERSALS.IN_ORDER(node.left, visitFn)
* Until it reach the bottom left node 'h' (b- d- h)
* h - visitFn get called
* h - TRAVERSALS.IN_ORDER(node.right, visitFn) get called
*
* d - visitFn get called
* d - left
* d - right
*
* b - visitFn
* b - left
* b - right
*/
IN_ORDER: (node, visitFn) => {
if (node !== null) {
console.log('left', node.left && node.left.key)
TRAVERSALS.IN_ORDER(node.left, visitFn)
console.log('call', node.key)
visitFn(node)
console.log('right', node.right && node.right.key)
TRAVERSALS.IN_ORDER(node.right, visitFn)
}
},
PRE_ORDER: (node, visitFn) => {
if (node !== null) {
visitFn(node)
TRAVERSALS.PRE_ORDER(node.left, visitFn)
TRAVERSALS.PRE_ORDER(node.right, visitFn)
}
},
POST_ORDER: (node, visitFn) => {
if (node !== null) {
TRAVERSALS.POST_ORDER(node.left, visitFn)
TRAVERSALS.POST_ORDER(node.right, visitFn)
visitFn(node)
}
}
} function createBinaryTree(rootKey) {
const root = createBinaryNode(rootKey) return {
root,
print(traversalType = 'IN_ORDER') {
let result = '' const visit = node => {
result += result.length === 0 ? node.key : ` => ${node.key}`
} TRAVERSALS[traversalType](this.root, visit) return result
}
}
} const tree = createBinaryTree('a')
const b = tree.root.addLeft('b')
const c = tree.root.addRight('c')
const d = b.addLeft('d')
const e = b.addRight('e')
const f = c.addLeft('f')
const g = c.addRight('g')
const h = d.addLeft('h')
const i = d.addRight('i') console.log('IN_ORDER: ', tree.print())// IN_ORDER: h => d => i => b => e => a => f => c => g
//console.log('PRE_ORDER: ', tree.print('PRE_ORDER')) // PRE_ORDER: a => b => d => h => i => e => c => f => g
  //console.log('POST_ORDER: ', tree.print('POST_ORDER')) // POST_ORDER: h => i => d => e => b => f => g => c => a
  exports.createBinaryNode = createBinaryNode
exports.createBinaryTree = createBinaryTree

Time complexity: O(n),

Space Complexity O(h) for average cases; h = logN -- this is because we need to stack all the function calls. worse cases: O(n)

[Algorithms] Build a Binary Tree in JavaScript and Several Traversal Algorithms的更多相关文章

  1. Construct Binary Tree from Inorder and Postorder Traversal

    Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder traversal of ...

  2. Construct Binary Tree from Preorder and Inorder Traversal

    Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a ...

  3. (二叉树 递归) leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal

    Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that ...

  4. (二叉树 递归) leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal

    Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that ...

  5. [LeetCode] Construct Binary Tree from Preorder and Inorder Traversal 由先序和中序遍历建立二叉树

    Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that ...

  6. 【LeetCode OJ】Construct Binary Tree from Preorder and Inorder Traversal

    Problem Link: https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-trave ...

  7. 36. Construct Binary Tree from Inorder and Postorder Traversal && Construct Binary Tree from Preorder and Inorder Traversal

    Construct Binary Tree from Inorder and Postorder Traversal OJ: https://oj.leetcode.com/problems/cons ...

  8. LeetCode:Construct Binary Tree from Inorder and Postorder Traversal,Construct Binary Tree from Preorder and Inorder Traversal

    LeetCode:Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder trav ...

  9. 【题解二连发】Construct Binary Tree from Inorder and Postorder Traversal & Construct Binary Tree from Preorder and Inorder Traversal

    LeetCode 原题链接 Construct Binary Tree from Inorder and Postorder Traversal - LeetCode Construct Binary ...

随机推荐

  1. OpenStack 计算服务 Nova计算节点部署 (九)

    如果使用vmware虚拟机进行部署,需要开启虚拟化:如果是服务器需要在bios上开启. Nova Compute nova-compute 一般运行在计算节点上,通过Messages Queue接收并 ...

  2. 9. Spark Streaming技术内幕 : Receiver在Driver的精妙实现全生命周期彻底研究和思考

        原创文章,转载请注明:转载自 听风居士博客(http://www.cnblogs.com/zhouyf/)       Spark streaming 程序需要不断接收新数据,然后进行业务逻辑 ...

  3. 深入解析php中的foreach问题

    本篇文章是对php中的foreach问题进行了详细的分析介绍,需要的朋友参考下   前言:php4中引入了foreach结构,这是一种遍历数组的简单方式.相比传统的for循环,foreach能够更加便 ...

  4. Codeforces Round #423 A Restaurant Tables(模拟)

    A. Restaurant Tables time limit per test 1 second memory limit per test 256 megabytes input standard ...

  5. 洛谷——P1014 Cantor表

    P1014 Cantor表 题目描述 现代数学的著名证明之一是Georg Cantor证明了有理数是可枚举的.他是用下面这一张表来证明这一命题的: 1/1 1/2 1/3 1/4 1/5 … 2/1 ...

  6. 2017广西邀请赛 Query on A Tree (可持续化字典树)

    Query on A Tree 时间限制: 8 Sec  内存限制: 512 MB提交: 15  解决: 3[提交][状态][讨论版] 题目描述 Monkey A lives on a tree. H ...

  7. Sqli-labs less 3

    Less-3 我们使用?id=' 注入代码后,我们得到像这样的一个错误: MySQL server version for the right syntax to use near "&qu ...

  8. Flask实战第50天:cms添加轮播图的模态对话框制作

    编辑cms_banners.html, 在{% block main_content%}中加上表给内容如下 {% block main_content %} <table class=" ...

  9. 通过myEclipse创建hibernate的实体类

    今天有个新项目中需要使用到hibernate,刚好数据库表已经创建完毕,就顺便来总结一下通过myEclipse创建hibernate的实体类. 1..在myEclipse中选择MyEclipse Da ...

  10. List the Books

    描述 Jim is fond of reading books, and he has so many books that sometimes it's hard for him to manage ...