Construct Binary Search Tree from Preorder Traversal.py

# 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.)

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

# Example 1:

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

 

# Constraints:

# 1 <= preorder.length <= 100
# 1 <= preorder[i] <= 10^8
# The values of preorder are distinct.



# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def bstFromPreorder(self, preorder: List[int]) -> TreeNode:

        def constructBST(preorder, start, end):
            if start > end:
                return None
            
            node = TreeNode(preorder[start])

            i = start
            while i <= end:
                if preorder[i] > node.val:
                    break
                i = i + 1

            node.left = constructBST(preorder, start + 1, i - 1)
            node.right = constructBST(preorder, i, end)

            return node
        
        return constructBST(preorder,0,len(preorder) - 1)