Solution.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, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def bstFromPreorder(self, preorder: List[int]) -> TreeNode:
        if not preorder:
            return None
        val = preorder[0]
        node = TreeNode(val)
        idx = 0
        for i, e in enumerate(preorder):
            if i == 0:
                continue
            if e < val:
                idx = i
            else:
                break
        left = self.bstFromPreorder(preorder[1:idx + 1])
        right = self.bstFromPreorder(preorder[idx + 1:])
        node.left = left
        node.right = right
        return node