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Node js path binary tree

how to return to a binary tree multiple nodes in a array or
a set
simple solution

There are a few issues in your code:
It expects internal nodes to have exactly two children. You should make use of the method getNumberOfChildren
path shouldn't be a global variable, as that will pollute any further call of getPaths. I would suggest to use a generator function, so that the caller can take care of putting the results in their own array variable.
Obviously root.getValue() === pathsToFind is never going to be true as pathsToFind is an array, and the root's value is an integer. You can use includes, but that will bring a bad time complexity when the array is long. So I'd suggest to turn that array into a Set.
It wasn't explained in the question, but it seems like you don't want to include a path to a node, when that node is on a path to another node that needs to be included. In other words, when the array of values has pairs where one is the ancestor of the other, the ancestor can be ignored.
Here is how I would suggest to implement it (with generator):
function* generatePaths(root, targetSet) {
if (!root) return;
const val = root.getValue();
let foundPathsWithThisValue = false;
for (let child = 0; child < root.getNumberOfChildren(); child++) {
for (const path of generatePaths(root.getChild(child), targetSet)) {
path.push(val);
yield path;
foundPathsWithThisValue = true;
}
}
if (!foundPathsWithThisValue && targetSet.has(val)) {
yield [val];
}
}
function getPaths(root, targetValues) {
return Array.from(generatePaths(root, new Set(targetValues)), path =>
path.reverse().join("->")
);
}
// No change below
class Node {
constructor(value) {
this.value = value;
this.children = [];
}
addChild(node) {
this.children.push(node);
}
getValue() {
return this.value;
}
getChild(i) {
return this.children[i];
}
getNumberOfChildren() {
return this.children.length;
}
print(indent = 0) {
console.log("-".repeat(indent), this.value);
for (const child of this.children) {
child.print(indent + 4);
}
}
}
const root = new Node(0);
const leftRoot = new Node(10);
const rightRoot = new Node(500);
root.addChild(leftRoot);
root.addChild(rightRoot);
const leftChild = new Node(150);
const rightChild = new Node(250);
leftRoot.addChild(leftChild);
leftRoot.addChild(rightChild);
const leftleftChild = new Node(200);
leftChild.addChild(leftleftChild);
const rightrightChild = new Node(300);
rightChild.addChild(rightrightChild);
// will output [ '0 -> 10 -> 150 -> 200', '0 -> 500' ]
console.log(getPaths(root, [150, 200, 500]));

You are free to use your accessors instead of root.children. But I guess it is clearer like this, at least for the answer.
Be careful, your property 'root' has the same name as the global variable root
function getPaths(root, pathsToFind) {
let match = false;
if (root === null) return [];
if (pathsToFind.includes(root.getValue())) {
match = true;
}
const pathChildren = root.children.map(child => getPaths(child, pathsToFind))
.filter(child => child.length > 0)
.map(value => `${root.getValue()} -- ${value}`);
if (pathChildren.length > 0) {
return pathChildren
} else {
return match ? [`${root.getValue()}`] : []
}
}

Why does this JavaScript SkipList implementation have "nodes" and "groups" which are structured the way they are?

A SkipList is a probabilistic data structure used, at least in part, for implementing an ordered key/value map. It is arranged in levels where higher levels skip nodes (the higher up, the more it skips), and the lowest level contains the nodes. As far as I have read, each level is implemented as a linked list, possibly a doubly linked list. And for some reason, skip lists are better for concurrency because in a multithreaded environment, they can be implemented without locks to enable fast/optimal performance as compared to say red/black trees or B+trees.
Here we have a "proper skip list" implemented in JavaScript, copied here for reference. The link has a nice suite of tests to show how it works.
const DEFAULT_STACK_UP_PROBABILITY = 0.25;
class ProperSkipList {
constructor(options) {
options = options || {};
this.stackUpProbability = options.stackUpProbability || DEFAULT_STACK_UP_PROBABILITY;
this.updateLength = options.updateLength !== false;
this.typePriorityMap = {
'undefined': 0,
'object': 1,
'number': 2,
'bigint': 2,
'string': 3
};
this.clear();
}
clear() {
let headNode = {
prev: null
};
let tailNode = {
next: null
};
this.head = {
isHead: true,
key: undefined,
value: undefined,
nodes: [headNode]
};
this.tail = {
isTail: true,
key: undefined,
value: undefined,
nodes: [tailNode]
};
headNode.next = tailNode;
tailNode.prev = headNode;
headNode.group = this.head;
tailNode.group = this.tail;
this.length = this.updateLength ? 0 : undefined;
}
upsert(key, value) {
let {matchingNode, prevNode, searchPath} = this._searchAndTrack(key);
if (matchingNode) {
let previousValue = matchingNode.group.value;
matchingNode.group.value = value;
return previousValue;
}
// Insert the entry.
let newNode = {
prev: prevNode,
next: prevNode.next
};
let newGroup = {
key,
value,
nodes: [newNode]
};
newNode.group = newGroup;
prevNode.next = newNode;
newNode.next.prev = newNode;
// Stack up the entry for fast search.
let layerIndex = 1;
while (Math.random() < this.stackUpProbability) {
let prevLayerNode = searchPath[layerIndex];
if (!prevLayerNode) {
let newHeadNode = {
prev: null,
group: this.head
};
let newTailNode = {
next: null,
group: this.tail
};
newHeadNode.next = newTailNode;
this.head.nodes.push(newHeadNode);
newTailNode.prev = newHeadNode;
this.tail.nodes.push(newTailNode);
prevLayerNode = newHeadNode;
}
let newNode = {
prev: prevLayerNode,
next: prevLayerNode.next,
group: newGroup
};
prevLayerNode.next = newNode;
newNode.next.prev = newNode;
newGroup.nodes.push(newNode);
layerIndex++;
}
if (this.updateLength) this.length++;
return undefined;
}
find(key) {
let {matchingNode} = this._search(key);
return matchingNode ? matchingNode.group.value : undefined;
}
has(key) {
return !!this.find(key);
}
_isAGreaterThanB(a, b) {
let typeA = typeof a;
let typeB = typeof b;
if (typeA === typeB) {
return a > b;
}
let typeAPriority = this.typePriorityMap[typeA];
let typeBPriority = this.typePriorityMap[typeB];
if (typeAPriority === typeBPriority) {
return a > b;
}
return typeAPriority > typeBPriority;
}
// The two search methods are similar but were separated for performance reasons.
_searchAndTrack(key) {
let layerCount = this.head.nodes.length;
let searchPath = new Array(layerCount);
let layerIndex = layerCount - 1;
let currentNode = this.head.nodes[layerIndex];
let prevNode = currentNode;
while (true) {
let currentNodeGroup = currentNode.group;
let currentKey = currentNodeGroup.key;
if (!currentNodeGroup.isTail) {
if (this._isAGreaterThanB(key, currentKey) || currentNodeGroup.isHead) {
prevNode = currentNode;
currentNode = currentNode.next;
continue;
}
if (key === currentKey) {
let matchingNode = currentNodeGroup.nodes[0];
searchPath[layerIndex] = matchingNode;
return {matchingNode, prevNode: matchingNode.prev, searchPath};
}
}
searchPath[layerIndex] = prevNode;
if (--layerIndex < 0) {
return {matchingNode: undefined, prevNode, searchPath};
}
currentNode = prevNode.group.nodes[layerIndex];
}
}
_search(key) {
let layerIndex = this.head.nodes.length - 1;
let currentNode = this.head.nodes[layerIndex];
let prevNode = currentNode;
while (true) {
let currentNodeGroup = currentNode.group;
let currentKey = currentNodeGroup.key;
if (!currentNodeGroup.isTail) {
if (this._isAGreaterThanB(key, currentKey) || currentNodeGroup.isHead) {
prevNode = currentNode;
currentNode = currentNode.next;
continue;
}
if (key === currentKey) {
let matchingNode = currentNodeGroup.nodes[0];
return {matchingNode, prevNode: matchingNode.prev};
}
}
if (--layerIndex < 0) {
return {matchingNode: undefined, prevNode};
}
currentNode = prevNode.group.nodes[layerIndex];
}
}
findEntriesFromMin() {
return this._createEntriesIteratorAsc(this.head.nodes[0].next);
}
findEntriesFromMax() {
return this._createEntriesIteratorDesc(this.tail.nodes[0].prev);
}
minEntry() {
let [key, value] = this.findEntriesFromMin().next().value;
return [key, value];
}
maxEntry() {
let [key, value] = this.findEntriesFromMax().next().value;
return [key, value];
}
minKey() {
return this.minEntry()[0];
}
maxKey() {
return this.maxEntry()[0];
}
minValue() {
return this.minEntry()[1];
}
maxValue() {
return this.maxEntry()[1];
}
_extractNode(matchingNode) {
let nodes = matchingNode.group.nodes;
for (let layerNode of nodes) {
let prevNode = layerNode.prev;
prevNode.next = layerNode.next;
prevNode.next.prev = prevNode;
}
if (this.updateLength) this.length--;
return matchingNode.group.value;
}
extract(key) {
let {matchingNode} = this._search(key);
if (matchingNode) {
return this._extractNode(matchingNode);
}
return undefined;
}
delete(key) {
return this.extract(key) !== undefined;
}
findEntries(fromKey) {
let {matchingNode, prevNode} = this._search(fromKey);
if (matchingNode) {
return {
matchingValue: matchingNode.group.value,
asc: this._createEntriesIteratorAsc(matchingNode),
desc: this._createEntriesIteratorDesc(matchingNode)
};
}
return {
matchingValue: undefined,
asc: this._createEntriesIteratorAsc(prevNode.next),
desc: this._createEntriesIteratorDesc(prevNode)
};
}
deleteRange(fromKey, toKey, deleteLeft, deleteRight) {
if (fromKey == null) {
fromKey = this.minKey();
deleteLeft = true;
}
if (toKey == null) {
toKey = this.maxKey();
deleteRight = true;
}
if (this._isAGreaterThanB(fromKey, toKey)) {
return;
}
let {prevNode: fromNode, searchPath: leftSearchPath, matchingNode: matchingLeftNode} = this._searchAndTrack(fromKey);
let {prevNode: toNode, searchPath: rightSearchPath, matchingNode: matchingRightNode} = this._searchAndTrack(toKey);
let leftNode = matchingLeftNode ? matchingLeftNode : fromNode;
let rightNode = matchingRightNode ? matchingRightNode : toNode.next;
if (leftNode === rightNode) {
if (deleteLeft) {
this._extractNode(leftNode);
}
return;
}
if (this.updateLength) {
let currentNode = leftNode;
while (currentNode && currentNode.next !== rightNode) {
this.length--;
currentNode = currentNode.next;
}
}
let leftGroupNodes = leftNode.group.nodes;
let rightGroupNodes = rightNode.group.nodes;
let layerCount = this.head.nodes.length;
for (let layerIndex = 0; layerIndex < layerCount; layerIndex++) {
let layerLeftNode = leftGroupNodes[layerIndex];
let layerRightNode = rightGroupNodes[layerIndex];
if (layerLeftNode && layerRightNode) {
layerLeftNode.next = layerRightNode;
layerRightNode.prev = layerLeftNode;
continue;
}
if (layerLeftNode) {
let layerRightmostNode = rightSearchPath[layerIndex];
if (!layerRightmostNode.group.isTail) {
layerRightmostNode = layerRightmostNode.next;
}
layerLeftNode.next = layerRightmostNode;
layerRightmostNode.prev = layerLeftNode;
continue;
}
if (layerRightNode) {
let layerLeftmostNode = leftSearchPath[layerIndex];
layerLeftmostNode.next = layerRightNode;
layerRightNode.prev = layerLeftmostNode;
continue;
}
// If neither left nor right nodes are present on the layer, connect based
// on search path to remove in-between entries.
let layerRightmostNode = rightSearchPath[layerIndex];
if (!layerRightmostNode.group.isTail) {
layerRightmostNode = layerRightmostNode.next;
}
let layerLeftmostNode = leftSearchPath[layerIndex];
layerLeftmostNode.next = layerRightmostNode;
layerRightmostNode.prev = layerLeftmostNode;
}
if (deleteLeft && matchingLeftNode) {
this._extractNode(matchingLeftNode);
}
if (deleteRight && matchingRightNode) {
this._extractNode(matchingRightNode);
}
}
_createEntriesIteratorAsc(currentNode) {
let i = 0;
return {
next: function () {
let currentGroup = currentNode.group;
if (currentGroup.isTail) {
return {
value: [currentNode.key, currentNode.value, i],
done: true
}
}
currentNode = currentNode.next;
return {
value: [currentGroup.key, currentGroup.value, i++],
done: currentGroup.isTail
};
},
[Symbol.iterator]: function () { return this; }
};
}
_createEntriesIteratorDesc(currentNode) {
let i = 0;
return {
next: function () {
let currentGroup = currentNode.group;
if (currentGroup.isHead) {
return {
value: [currentNode.key, currentNode.value, i],
done: true
}
}
currentNode = currentNode.prev;
return {
value: [currentGroup.key, currentGroup.value, i++],
done: currentGroup.isHead
};
},
[Symbol.iterator]: function () { return this; }
};
}
}
Where I am coming from is wondering how the theoretical concepts of a SkipList are applied in this implementation. But that is not my question, my question is very specific as described at the bottom. But take a look at a few things first. For example, the clear function is implemented like this, creating a fresh new SkipList:
clear() {
let headNode = {
prev: null
};
let tailNode = {
next: null
};
this.head = {
isHead: true,
key: undefined,
value: undefined,
nodes: [headNode]
};
this.tail = {
isTail: true,
key: undefined,
value: undefined,
nodes: [tailNode]
};
headNode.next = tailNode;
tailNode.prev = headNode;
headNode.group = this.head;
tailNode.group = this.tail;
this.length = this.updateLength ? 0 : undefined;
}
Notice the "group" as in headNode.group = this.head has this structure:
{
isTail: boolean,
isHead: boolean.
key: undefined,
value: undefined,
nodes: [node]
}
Or inserting a new node:
// Insert the entry.
let newNode = {
prev: prevNode,
next: prevNode.next
};
let newGroup = {
key,
value,
nodes: [newNode]
};
newNode.group = newGroup;
prevNode.next = newNode;
newNode.next.prev = newNode;
So a node has a prev/next/group set of properties, while a group has a key/value/nodes. Finally, notice that this.head.nodes[layerIndex] and such is being called in several places, which to me seems like it simply bypasses the linked-list nature of the SkipList and has an internal array that stores all the nodes anyway. Basically I am confused, I have been looking at this code for a day and it looks to me like each level node is storing an array of all the children nodes for that level, or something like that. I am confused with what is happening here.
Can one explain what the general model is for this SkipList implementation? That is:
The meaning of group and node objects, and why they have their structure they do.
What the nodes array is doing on a group.
For context, I am trying to learn how to implement a SkipList, after reading several papers on it and understanding the general idea. I want to put some constraints on it, so don't want to use any existing implementation. I want it to work in JavaScript and C, so it needs to be single-threaded in one context and multithreaded in the other, so I have a bit to learn. But knowing how to implement it in JavaScript, like what the above code is doing, is a good first step.

The summary of your question is:
The meaning of group and node objects, and why they have their structure they do.
What the nodes array is doing on a group.
I'll use an image taken from brilliant.org:
The linked lists appear as horizontal bands, while the arrays appear as vertical stacks.
A group corresponds to one distinct key in the set -- marked in yellow. The corresponding value is not depicted in the image, but is just payload and is not essential for understanding the data structure. A group has a nodes array -- displayed as a stack on top of the key. Each of these nodes belong to a different linked list in which they represent that same key. Those nodes have backreferences again to the group, so a node knows which key it represents, and which other linked lists have a node for this same key.
So the nodes array actually duplicates a key to different lists. One nodes array does not represent different keys in the collection -- just one. It is like an elevator to jump from one train (linked list) to another, with the same key.
The sizes of these arrays are randomly determined, but are intended to be small on average. This is driven by the stackUpProbability. It represents the probability that an array is extended with one more node (upwards in the image). The probability for having at least one node (for a key) is obviously 1, but the probability that the array will have at least one more node in it (and thus have a size of at least 2), is stackUpProbability. The probability that it will have size 3 is (stackUpProbability)²... etc.
This way, the bottom linked list (at layerIndex 1) will have all keys in the collection (one node per key, in sorted order), but any next layer will have fewer keys, and the top layer might only have a hand full. The lists with fewer nodes provide a way to travel fast along the collection in search of a value. This provides a proximity of where a value could be found. Via the "elevator" a search algorithm can step down to lower (slower) linked lists and continue the search there, until it reaches the lowest level where the key will be found (if present), or where it would have been (if not).

Print a hierarchical tree structure in javascript

I'm working on a problem where given an array of file paths I would like to print the file structure. For example with the given array ["/a/b/c", "a/a/a", "/a/b/d"], the ideal structure would look like :
a
b
c
d
a
a
But my structure ends up looking more like this:
a
b
c
a
a
a
b
From what I can gather this is being caused by my tree not recognizing when a node already exists in a tree. So it is adding the node "a" three times as opposed to recognizing that an "a" already exists and traversing into it.
let paths = ["/a/b/c", "a/a/a", "/a/b/d"]
class TreeNode {
constructor(value) {
this.value = value;
this.children = [];
}
addChild(element) {
this.children.push(element)
}
}
const head = new TreeNode('Head');
let cur = head;
paths.forEach(element => {
cur = head;
let filePath = element.split('/');
filePath.shift();
filePath.forEach(element => {
let newNode = new TreeNode(element);
if(!cur.children.includes(newNode)) {
cur.addChild(newNode);
cur = cur.children[cur.children.length - 1];
} else {
cur = cur.children.indexOf(newNode);
}
})
})
var spaceAppend = function(num) {
let i = 0;
let space = "";
while(i < num) {
space += " ";
i++;
}
return space;
}
var traverse = function(node, level = 0){
if(node === null)
return;
console.log(spaceAppend(level), node.value)
if(node.children) {
for(const n of node.children) {
traverse(n, level + 1);
}
}
}
traverse(head)
Is there an issue with my tree implementation?

Some issues:
.includes() is not the right way to find a matching value. Use .find() instead.
.indexOf() will return an index, so that is not the right value you want to assign to cur in the else block.
shift may throw away an essential part of the path when it does not start with /. You can ease the processing by using .match() instead of .split(), so that you get exactly the non-empty parts of the path.
Less of an issue:
There is no need to define cur outside of the outer loop.
JavaScript has a native function for something like spaceAppend. You can use .repeat().
new TreeNode(element) is also called when you actually don't need it. Only create a new node when you know there is no matching node.
You could replace the inner .forEach() loop with .reduce(), which gives a better scope-handling for the cur variable.
Here is your code with those remarks taken into account:
class TreeNode {
constructor(value) {
this.value = value;
this.children = [];
}
addChild(element) {
this.children.push(element);
}
}
let paths = ["/a/b/c", "a/a/a", "/a/b/d"];
const head = new TreeNode('Head');
paths.forEach(element => {
// Use .match() to only get non-empty elements
let filePath = element.match(/[^\/]+/g);
filePath.reduce((cur, element) => {
// Use .find() instead of .includes()
let node = cur.children.find(child => child.value === element);
// Only create the node when needed:
if (!node) {
node = new TreeNode(element);
cur.addChild(node);
}
// Walk down one step in the tree
return node; // ...becomes the value of `cur`
}, head); // Initial value of reduction
});
const traverse = function(node, level=0) {
if (node === null) return;
// Use .repeat():
console.log(" ".repeat(level), node.value);
if (node.children) {
for (const n of node.children) {
traverse(n, level + 1);
}
}
}
traverse(head);

Is the starter array meant to be ["/a/b/c", "/a/a/a", "/a/b/d"] ("/a/a/a" instead of ("a/a/a")?
I think the crux of the problem you're having is the line
if(!cur.children.includes(newNode)) { ... }
When a new node is created, even if it has the same value as a previous one, it will not result in equity when comparing the two TreeNode objects. You need to compare the value of the nodes, not the nodes themselves.
So an example with a simplified version of your node object:
class TreeNode {
constructor(value) {
this.value = value;
}
}
a1 = new TreeNode('a');
a2 = new TreeNode('a');
console.log("a1 == a2");
console.log(a1 == a2); // false
console.log("a1.value == a2.value");
console.log(a1.value == a2.value); // true
I adjusted the inner forEach loop with one that compares the values instead of the TreeNode objects
filePath.forEach(element => {
let newNode = new TreeNode(element);
let tempNode = null;
for (var i = 0; i < cur.children.length; i++) {
if (cur.children[i].value == newNode.value) {
tempNode = cur.children[i];
}
}
if (tempNode == null) {
cur.addChild(newNode);
cur = newNode;
} else {
cur = tempNode;
}
});
Full code snippet on codepen
Object equality in javascript isn't particularly nice to deal with see this other answer for more information

Here is a solution using lodash and object-treeify. While it's simpler code, there is obviously a trade-off introducing additional dependencies.
This solution works by first converting the paths into a tree structure and then visualizing it using object-treeify
// const lodash = require('lodash');
// const objectTreeify = require('object-treeify');
const myPaths = ['/a/b/c', 'a/a/a', '/a/b/d'];
const treeify = (paths) => objectTreeify(paths.reduce((p, c) => {
lodash.set(p, c.match(/[^/]+/g));
return p;
}, {}), {
spacerNoNeighbour: ' ',
spacerNeighbour: ' ',
keyNoNeighbour: '',
keyNeighbour: ''
});
console.log(treeify(myPaths));
/* =>
a
b
c
d
a
a
*/
.as-console-wrapper {max-height: 100% !important; top: 0}
<script src="https://bundle.run/lodash#4.17.20"></script>
<script src="https://bundle.run/object-treeify#1.1.31"></script>
Disclaimer: I'm the author of object-treeify

Recursively concatenate passed string

I'm writing a BST in JS and trying to concatenate a string passed into a member function without success. console.log() works but it automatically starts a new line for each entry.
"use strict";
class Node {
constructor(dt) {
this.data = dt;
this.left = null;
this.right = null;
}
}
class BST {
constructor() {
this.root = null;
}
insert(parent, key) {
if (!this.root) return this.root = new Node(key);
else {
if (key < parent.data) {
if (!parent.left) {
parent.left = new Node(key);
} else {
this.insert(parent.left, key);
}
} else if (key > parent.data) {
if (!parent.right) {
parent.right = new Node(key);
} else {
this.insert(parent.right, key);
}
}
}
}
// Doesn't work
printIN(parent, string) {
if (parent) {
this.printIN(parent.left);
console.log(parent.data);
string += " " + parent.data;
// return string += " " + parent.data;
this.printIN(parent.right);
}
return string;
}
// This works.
// printIN(parent) {
// if (parent) {
// this.printIN(parent.left);
// console.log(parent.data);
// this.printIN(parent.right);
// }
// }
}
let treeA = new BST();
let tree = null;
tree = treeA.insert(treeA.root, 5);
tree = treeA.insert(treeA.root, 7);
tree = treeA.insert(treeA.root, 3);
tree = treeA.insert(treeA.root, 14);
let string = [""];
string = treeA.printIN(treeA.root, string);
console.log();
console.log(string);
// treeA.printIN(treeA.root);
I want to print out the numbers on one single line, instead of them starting on a new line each time. I thought using string concatenation is the logical solution, but I can't seem to make it work.
// Doesn't work
printIN(parent, string) {
if (parent) {
this.printIN(parent.left);
console.log(parent.data);
string += " " + parent.data;
// return string += " " + parent.data;
this.printIN(parent.right);
}
return string;

Another answer gave you a solution to your string problem. The snippet below has an alternative. But more importantly, it offers some significant clean-up to your OOP approach.
There are two main reasons this is necessary.
First, you have member functions of your object, which still require the object to be passed to them. That's redundant and confusing. Any OOP function which doesn't make reference to this should be a static function. But if you're going to do OOP (I might argue against it in fact), then you shouldn't use static functions as though they were methods. Thus insert and printIN should not have the parent parameter.
Second, you are dealing with a recursive structure. Nodes of a tree should have the same design as the tree itself. This makes all sorts of things easier. But you have a class Node and one called BST. And BST then has a root node, which should then be recursive. There is simply no reason for this that I can see. A BST by nature is a root node, with a data value (usually, although we can make that optional as you do) and potentially left and right nodes that are themselves BSTs.
Fixing the code to work like this simplifies it substantially. Here is one implementation:
class BST {
constructor(data) {
this.data = data;
}
insert(key) {
if (!this.data) {
this.data = key;
} else {
if (key < this.data) {
if (!this.left) {
this.left = new BST(key);
} else {
this.left.insert(key);
}
} else if (key > this.data) {
if (!this.right) {
this.right = new BST(key);
} else {
this.right.insert(key);
}
}
}
}
printIN() {
return (this.left ? this.left.printIN() + ' ' : '') +
this.data +
(this.right ? ' ' + this.right.printIN() : '');
}
}
let treeA = new BST();
treeA.insert(5);
treeA.insert(7);
treeA.insert(3);
treeA.insert(14);
console.log(treeA.printIN());
There is no Node class here. Everything is done in BST. Moreover, BST objects don't need a root node and the methods are simpler.
I would note, though that printIN is less useful than another function which we could use the same way:
toArray() {
return [
...(this.left ? this.left.toArray() : []),
this.data,
...(this.right ? this.right.toArray() : [])
];
}
which would allow
console.log(treeA.toArray()); //=> [3, 5, 7, 14]
and you could get your string just by joining that array with a space. This function strikes me as much more useful than the printIN one you have.
We could also add a static function:
static fromArray(xs) {
return xs .reduce (function (tree, x) {
tree.insert(x)
return tree
}, new BST())
}
that would allow us to write
tree = BST .fromArray ([8, 6, 7, 5, 3, 0, 9]);
tree .printIN() //=> [0, 3, 5, 6, 7, 8, 9]
I mentioned that I wouldn't even do this with OOP. But I don't do much that way these days, preferring functional approaches. So, if you're curious, here's another take on the problem altogether, using just functions, and trees that aren't modified when you do an insert, but instead return new trees:
const bst = (xs) =>
xs .reduce ((tree, x) => bstInsert (x, tree), {})
const bstInsert = (x, {left, data, right} = {}) =>
data == undefined
? {data: x}
: x < data
? {data, left: bstInsert (x, left), ...(right ? {right} : {})}
: {data, ...(left ? {left} : {}), right: bstInsert (x, right)}
const inorder = ({left, data, right} = {}) => [
...(left ? inorder(left) : []),
data,
...(right ? inorder(right) : [])
]
inorder (bst ([8, 6, 7, 5, 3, 0, 9])) //=> [0, 3, 5, 6, 7, 8, 9]
I'm not recommending that technique, specifically, but FP can be a powerful alternative to OOP.

Try this technique.
printIN(parent, result = []) {
if (parent) {
this.printIN(parent.left, result);
result.push(parent.data)
this.printIN(parent.right, result);
}
return result;
}
Pass, as a second argument, the array of strings that will serve as the resulting list of node values as per the in-order bst traversal.
Use a default value for this argument of [] then, if omitted, one will be created, but when recursing, you can pass the existing array in. When you visit a node, push the value into the array. Call printIN on the root node and do not provide the result argument (a new array will be created and returned).
If you want to log all on one line, use Array.prototype.join to transform the array of strings into a single comma-separated string, and log that.
Working example:
class Node {
constructor(dt) {
this.data = dt;
this.left = null;
this.right = null;
}
}
class BST {
constructor() {
this.root = null;
}
insert(parent, key) {
if (!this.root) return this.root = new Node(key);
else {
if (key < parent.data) {
if (!parent.left) {
parent.left = new Node(key);
} else {
this.insert(parent.left, key);
}
} else if (key > parent.data) {
if (!parent.right) {
parent.right = new Node(key);
} else {
this.insert(parent.right, key);
}
}
}
}
printIN(parent, result = []) {
if (parent) {
this.printIN(parent.left, result);
result.push(parent.data)
this.printIN(parent.right, result);
}
return result;
}
}
let treeA = new BST();
let tree = null;
tree = treeA.insert(treeA.root, 5);
tree = treeA.insert(treeA.root, 7);
tree = treeA.insert(treeA.root, 3);
tree = treeA.insert(treeA.root, 14);
console.log(treeA.printIN(treeA.root).join());

create children properties recursively in Javascript

I am trying to create a depth of children based on a passed parameter n.
Basically if n is 4, the resulting object should be
parent.children.children.children.children.
I've come up with this so far:
parent = {}
function makechildren( current, depth ){
current['children']={}
while (depth>0){ {
str = JSON.stringify(current)
return makechildren(current, depth-1)
}
}
}
makechildren(parent, 4)

I tested this code and it works
parent={};
var obj;
function makechildren( current, depth){
if(depth>0)
{
current = JSON.parse(JSON.stringify(current).replace('{}','{"children":{}}'))
makechildren(current, depth-1);
}else{
obj = current;
return ;
}
}
console.log(obj)