Construct Binary Search Tree from Preorder Traversal

Problem is here: https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/

Return the root node of a binary search tree that matches the given preorder traversal.

(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val.  Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)

Example 1:

Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]

Note: 

1. 1 <= preorder.length <= 100
2. The values of preorder are distinct.

This is a relatively simple problem, although labeled as Medium. The definition of a binary search tree is even given in the problem description. A recursive solution does the trick: add the first node, then call recursively for the left by creating a sub-array with the smaller elements, and similar to the right. That should do it. Conversion of linked-lists to arrays in C# is not the most efficient operation, hence the execution time slows down a bit. Code is below, cheers, ACC.

public class Solution
{
 public TreeNode BstFromPreorder(int[] preorder)
 {
  if (preorder == null || preorder.Length == 0) return null;
  TreeNode tree = new TreeNode(preorder[0]);
  LinkedList left = new LinkedList();
  int i = 1;
  for (; i < preorder.Length && preorder[i] < preorder[0]; i++)
  {
   left.AddLast(preorder[i]);
  }
  LinkedList right = new LinkedList();
  for (; i < preorder.Length; i++)
  {
   right.AddLast(preorder[i]);
  }
  tree.left = BstFromPreorder(left.ToArray());
  tree.right = BstFromPreorder(right.ToArray());
  return tree;
 }
}