# 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] <= 108
# All the values of preorder are unique.

# Definition for a binary tree node.


class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


class Solution(object):
    def bstFromPreorder(self, preorder):

        def build(left, right):
            if left > right:
                return None
            node = TreeNode(preorder[left])

            if left == right:
                return node

            index = left + 1

            while index <= right and preorder[left] > preorder[index]:
                index += 1

            node.left = build(left + 1, index - 1)
            node.right = build(index, right)

            return node

        return build(0, len(preorder) - 1)
