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1008. 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.)

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.

Solutions (Python)

1. Recursion

# 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:
        if not preorder:
            return None

        return TreeNode(
            preorder[0],
            self.bstFromPreorder(
                [val for val in preorder if val < preorder[0]]),
            self.bstFromPreorder(
                [val for val in preorder if val > preorder[0]])
        )