Leetcode #255: Verify Preorder Sequence in Binary Search Tree
In this guide, we solve Leetcode #255 Verify Preorder Sequence in Binary Search Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an array of unique integers preorder, return true if it is the correct preorder traversal sequence of a binary search tree. Example 1: Input: preorder = [5,2,1,3,6] Output: true Example 2: Input: preorder = [5,2,6,1,3] Output: false Constraints: 1 <= preorder.length <= 104 1 <= preorder[i] <= 104 All the elements of preorder are unique.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Stack, Tree, Binary Search Tree, Recursion, Array, Binary Tree, Monotonic Stack
Intuition
We need the next greater or smaller element efficiently, which is exactly what a monotonic stack offers.
Each element is pushed and popped at most once, yielding a linear-time scan.
Approach
Maintain a stack that is either increasing or decreasing, depending on the query.
When the invariant is broken, pop and resolve answers for those indices.
Steps:
- Scan elements once.
- Pop while the monotonic condition is violated.
- Use stack indices to update answers.
Example
Input: preorder = [5,2,1,3,6]
Output: true
Python Solution
class Solution:
def verifyPreorder(self, preorder: List[int]) -> bool:
stk = []
last = -inf
for x in preorder:
if x < last:
return False
while stk and stk[-1] < x:
last = stk.pop()
stk.append(x)
return True
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.