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)
};