day_20.py

'''
Problem Statement:
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.
'''

# Solution:

# 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
        root = TreeNode(preorder[0])
        i = 1
        while i < len(preorder):
            if preorder[i] < preorder[0]: i += 1
            else: break
        root.left = self.bstFromPreorder(preorder[1:i])
        root.right = self.bstFromPreorder(preorder[i:])
        return root