# 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) };