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1008. Construct Binary Search Tree from Preorder Traversal

Difficulty Topics

Description

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] <= 1000
  • All the values of preorder are unique.

Solution

construct-binary-search-tree-from-preorder-traversal.py
# 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]) -> Optional[TreeNode]:

        def go(A, bound = float('inf')):
            if not A or A[-1] > bound: return None

            node = TreeNode(A.pop())
            node.left = go(A, node.val)
            node.right = go(A, bound)

            return node

        return go(preorder[::-1])