How to make binary tree from array in javascript?

Question

I have an array var array = [8,10,12,5,3,6];

Logic

1. First node would be root node.
2. If new node value is less or equal =< than parent node, It would be left node of parent node
3. If new node value is greater > than parent node, It would be right node of parent node

And I am trying to achieve output like below object:

{
   value:8,
   left:{
      value:5,
      left:{ value:3 },
      right:{value:6}
   },
   right:{
      value:10,
      right:{value:12}
   }
}

Which would be in image like this

I tried below code:

var arr = [8,10,12,5,3,6];
var root = arr[0];
var rv = {};
for (var i = 0; i < arr.length; i++){
    if(arr[i] < root){
    rv.left = arr[i];
  }else{
    rv.right = arr[i];
  }
}
console.log(rv);

Please help me to solve this.

Answer 1

You could use a Node instance for new nodes and a function for inserting nodes.

Then iterate the values and build a new tree.

function Node(value) {
    this.value = value;
    // this.left = null;
    // this.right = null;
}

function insertNode(tree, value) {
    var node = tree,
        key;
    while (node.value !== value) {
         key = value < node.value ? 'left' : 'right';
         if (!node[key]) {
             node[key] = new Node(value);
             break;
         }
         node = node[key];
    }
    return tree;
}

var array = [8, 10, 12, 5, 3, 6],
    tree = array.reduce((t, v) => t ? insertNode(t, v) : new Node(v), null);

console.log(tree);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

While it is close to Nina's answer i believe to be a little more concise;

var data = [8,10,12,5,3,6],
    tree;

function insertBinTree (t = {value: void 0, left: void 0, right: void 0}, n){
  t.value !== void 0 ? t.value > n ? t.left = insertBinTree(t.left,n)
                                   : t.right = insertBinTree(t.right,n)
                     : t.value = n;
  return t;
}

tree = data.reduce(insertBinTree, void 0);
console.log(tree);

.as-console-wrapper {
max-height: 100% !important
}

Answer 3

Try something like this by recursive method

var binary = {};
var arr = [8,5,10,3,6,12];

function makeBinary(binary,number){
  if(binary.value === undefined){
    binary.value = number;
  }else if(number > binary.value){
    if(binary.right === undefined){
      binary.right = {value:number};  
    }else{
      binary.right = makeBinary(binary.right,number);
    }
  }else{
    if(binary.left === undefined){
      binary.left = {value:number};  
    }else{
      binary.left = makeBinary(binary.left,number);
    }
  }
  return binary;
}

for(let i in arr){
  makeBinary(binary,arr[i]);
}

console.log(binary);

Answer 4

class Node {
    constructor(val){
        this.val=val;
        this.right=null;
        this.left=null;
    }
}


class Bst{
    constructor(){
        this.root=null;
    }

    insert(val){
        let newNode= new Node (val)
        if (!this.root) this.root=newNode;
    let current =this.root;        
        while (true) {
            if(val === current.val) return undefined;
            if(current.val<val){
                if (current.right===null){
                    current.right=newNode;
                    return this
                }
                    else
                    current=current.right}
            if(current.val>val){
                if (current.left===null){
                    current.left=newNode;
                    return this
                }
                    else
                    current=current.left}
                
        }
        
    
    }
    print (){
        let all="Root=";
    let visit=(current=this.root)=>{
        if(!current.left && !current.right){
            if(all[all.length-1]<current.val)
                 all+=`,LeR${current.val}`
                 else
                 all+=`,LeL${current.val}`
                }
else{
    if(all[all.length-1]<current.val)
         all+=`,FR${current.val}`
         else
                  all+=`,FL${current.val}`
            }
       
   if (current.left) visit(current.left)
   if (current.right)  visit(current.right)
        }
        visit()
        all+=` ,valid bst:${this.isValidBST()}`
        return all

    }

 isValidBST(node=this.root, min = null, max = null) {
if (!node) return true;
if (max !== null && node.data >= max) {
  return false;
}
if (min !== null && node.data <= min) {
  return false;
}
const leftSide = this.isValidBST(node.left, min, node.data);
const rightSide = this.isValidBST(node.right, node.val, max);

return leftSide && rightSide;
    }

find(val){
    let found=false
    let innerFind=(current=this.root)=>{
    if (val>current.val) 
       if (current.right != null) return innerFind(current.right)
        if(val===current.val)
        found= true 
   else
      if (current.left != null)return innerFind(current.left)
        if(val===current.val)
            found= true 
    return found
        }
        return innerFind()
    }

}




let tree=new Bst

tree.insert(8)
tree.insert(10)
tree.insert(13)
tree.insert(3)
tree.insert(1)
tree.insert(6)
tree.insert(4)
tree.insert(7)
tree.insert(2)
tree.print()

Answer 5

You can do this by technique is called recursion. make an array with the structure ( left_subtree, key, right_subtree) in your case var array = [[3,5,6],8,[null,10,12]

class TreeNode {
    constructor(key) {
        this.key = key;
        this.right = null;
        this.left = null;
    }
}

function parseTuple(data) {
    if (data === null) {
        let node = null;
        return node;
    }
    else if (data.length === 3) {
        let node = new TreeNode(data[1]);
        node.left = parseTuple(data[0]);
        node.right = parseTuple(data[2]);
        return node;
    }
    else {
        let node = new TreeNode(data);
        return node;
    }

}