LeetCode
LeetCode 1008 Construct Binary Search Tree From Preorder Traversal - Medium
1008. Construct Binary Search Tree from Preorder Traversal -- Medium
1008. Construct Binary Search Tree From Preorder Traversal — Medium
Problem
- Construct Binary Search Tree from Preorder Traversal -- Medium
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, 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 helper(lower = float('-inf'), upper = float('inf')):
nonlocal idx
# if all elements from preorder are used
if idx == n:
return None
val = preorder[idx]
# if the current element not meet BST requirements
if val < lower or val > upper:
return None
# place the current element
idx += 1
root = TreeNode(val)
root.left = helper(lower, val)
root.right = helper(val, upper)
return root
idx = 0
n = len(preorder)
return helper()