# Construct Binary Search Tree from Preorder Traversal

# 1008. Construct Binary Search Tree from Preorder Traversal

Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.

It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.

A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val.

A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.

```### Example 1:
```
Input: preorder = [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]
```
```
```### Example 2:
```
Input: preorder = [1,3]
Output: [1,null,3]
```
```
```Constraints:
```
1 <= preorder.length <= 100
1 <= preorder[i] <= 1000
All the values of preorder are unique.
```
```
```/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {number[]} preorder
* @return {TreeNode}
*/
var bstFromPreorder = function(preorder) {
const makeBST = (node) => {
if(!node.length) {
return null;
}

let root = new TreeNode(node[0]);
root.left = bstFromPreorder(node.filter(el => el < node[0]));
root.right = bstFromPreorder(node.filter(el => el > node[0]));
return root;
}

return makeBST(preorder)
};
```