Sync LeetCode submission Runtime - 15 ms (42.30%), Memory - 18.7 MB (76.51%)

Author: jimit105

0255-verify-preorder-sequence-in-binary-search-tree/README.md

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+<p>Given an array of <strong>unique</strong> integers <code>preorder</code>, return <code>true</code> <em>if it is the correct preorder traversal sequence of a binary search tree</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/12/preorder-tree.jpg" style="width: 292px; height: 302px;" />
+<pre>
+<strong>Input:</strong> preorder = [5,2,1,3,6]
+<strong>Output:</strong> true
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> preorder = [5,2,6,1,3]
+<strong>Output:</strong> false
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= preorder.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= preorder[i] &lt;= 10<sup>4</sup></code></li>
+	<li>All the elements of <code>preorder</code> are <strong>unique</strong>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong> Could you do it using only constant space complexity?</p>

0255-verify-preorder-sequence-in-binary-search-tree/solution.py

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+# Approach 1: Monotonic Stack
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def verifyPreorder(self, preorder: List[int]) -> bool:
+        min_limit = -inf
+        stack = []
+
+        for num in preorder:
+            while stack and stack[-1] < num:
+                min_limit = stack.pop()
+
+            if num <= min_limit:
+                return False
+
+            stack.append(num)
+
+        return True