BST with Dead End | GfG POTD | 09-06-2025 | GfG Problem of the day | GeeksforGeeks
🎯 In this video, we solve the “BST with Dead End” problem from GeeksforGeeks. Given a Binary Search Tree (BST) containing unique positive integers (values greater than 0), determine whether it contains a “dead end”—a leaf node where no new node can be inserted while preserving BST properties and the constraint that all values remain positive.
🔗 Problem Reference →
Check Whether BST Contains Dead End → https://www.geeksforgeeks.org/problems/check-whether-bst-contains-dead-end/1
🧾 Problem Statement
You’re given a BST of unique positive integers. A dead end is a leaf node such that there is no valid integer you could insert as a descendant of that node (because its allowable range spans exactly one value). Return true if at least one dead end exists, otherwise false.
📊 Examples:
• Input: root = [8, 5, 9, 2, 7, N, N, 1]
Output: true
Explanation: Node 1 has bounds (min = 0, max = 2), so no integer between 0 and 2 other than 1 itself can be inserted.
• Input: root = [8, 7, 10, 2, N, 9, 13]
Output: true
Explanation: Node 9 has bounds (min = 8, max = 10), so it’s a dead end.
🧠 Concepts & Techniques Covered
🔹 Approach (DFS with Min/Max Bounds)
– Recursively traverse the BST, passing down min and max for each subtree.
– At a leaf, check if node→data – min == 1 and max – node→data == 1. If so, it’s a dead end.
🔹 Time Complexity → O(N) — each node is visited once.
🔹 Space Complexity → O(H) — recursion stack of height H (O(log N) balanced, O(N) skewed).
💡 Video Breakdown
00:00 → Introduction & Problem Overview
00:40 → Understanding Dead End Concept
07:08 → DFS Min/Max Bounds Intuition
12:38 → Code
📂 Code & Resources
• GeeksforGeeks Problem Link → https://www.geeksforgeeks.org/problems/check-whether-bst-contains-dead-end/1
• GitHub Link → https://github.com/imnilesh18/GfG-POTD/blob/master/06_June/09_BST%20with%20Dead%20End.cpp
• GitHub Repo (All POTD Solutions) → https://github.com/imnilesh18/GfG-POTD
👨💻 What You’ll Learn
✅ How to leverage DFS with dynamic min/max bounds in a BST
✅ Detecting insertion constraints at leaf nodes
✅ Translating the approach into clean C++ and Java code
✅ Handling edge cases (e.g., value = 1 at the lower bound)
💬 If this helped you, Like 👍, Comment 💬 your questions or improvements, and Subscribe 🔔 to LeetGeek for concise, example-driven DSA tutorials with full dry runs and code breakdowns!
🔖 Tags
#GFG #BSTDeadEnd #BinarySearchTree #DFS #Recursion #LeetGeek #GFGPOTD #DailyCodingChallenge #DSA #Cplusplus #Java #InterviewPrep #GeekStreak2025 #CodingTutorial #ProblemSolving #Algorithm #CodingInterviewPrep #ProgrammingTutorial #SoftwareEngineering #TechInterviewQuestions #DataStructuresAndAlgorithms #AlgorithmTutorial #TreeAlgorithms #LeetCodePractice #CompetitiveProgramming #CodeWithMe #DeveloperLife #100DaysOfCode #CodeDaily #ProgrammingCommunity #TechEducation #YouTubeLearning #Shorts #ProblemSolving #CodeReview #DebuggingTips
Видео BST with Dead End | GfG POTD | 09-06-2025 | GfG Problem of the day | GeeksforGeeks канала LeetGeek
🔗 Problem Reference →
Check Whether BST Contains Dead End → https://www.geeksforgeeks.org/problems/check-whether-bst-contains-dead-end/1
🧾 Problem Statement
You’re given a BST of unique positive integers. A dead end is a leaf node such that there is no valid integer you could insert as a descendant of that node (because its allowable range spans exactly one value). Return true if at least one dead end exists, otherwise false.
📊 Examples:
• Input: root = [8, 5, 9, 2, 7, N, N, 1]
Output: true
Explanation: Node 1 has bounds (min = 0, max = 2), so no integer between 0 and 2 other than 1 itself can be inserted.
• Input: root = [8, 7, 10, 2, N, 9, 13]
Output: true
Explanation: Node 9 has bounds (min = 8, max = 10), so it’s a dead end.
🧠 Concepts & Techniques Covered
🔹 Approach (DFS with Min/Max Bounds)
– Recursively traverse the BST, passing down min and max for each subtree.
– At a leaf, check if node→data – min == 1 and max – node→data == 1. If so, it’s a dead end.
🔹 Time Complexity → O(N) — each node is visited once.
🔹 Space Complexity → O(H) — recursion stack of height H (O(log N) balanced, O(N) skewed).
💡 Video Breakdown
00:00 → Introduction & Problem Overview
00:40 → Understanding Dead End Concept
07:08 → DFS Min/Max Bounds Intuition
12:38 → Code
📂 Code & Resources
• GeeksforGeeks Problem Link → https://www.geeksforgeeks.org/problems/check-whether-bst-contains-dead-end/1
• GitHub Link → https://github.com/imnilesh18/GfG-POTD/blob/master/06_June/09_BST%20with%20Dead%20End.cpp
• GitHub Repo (All POTD Solutions) → https://github.com/imnilesh18/GfG-POTD
👨💻 What You’ll Learn
✅ How to leverage DFS with dynamic min/max bounds in a BST
✅ Detecting insertion constraints at leaf nodes
✅ Translating the approach into clean C++ and Java code
✅ Handling edge cases (e.g., value = 1 at the lower bound)
💬 If this helped you, Like 👍, Comment 💬 your questions or improvements, and Subscribe 🔔 to LeetGeek for concise, example-driven DSA tutorials with full dry runs and code breakdowns!
🔖 Tags
#GFG #BSTDeadEnd #BinarySearchTree #DFS #Recursion #LeetGeek #GFGPOTD #DailyCodingChallenge #DSA #Cplusplus #Java #InterviewPrep #GeekStreak2025 #CodingTutorial #ProblemSolving #Algorithm #CodingInterviewPrep #ProgrammingTutorial #SoftwareEngineering #TechInterviewQuestions #DataStructuresAndAlgorithms #AlgorithmTutorial #TreeAlgorithms #LeetCodePractice #CompetitiveProgramming #CodeWithMe #DeveloperLife #100DaysOfCode #CodeDaily #ProgrammingCommunity #TechEducation #YouTubeLearning #Shorts #ProblemSolving #CodeReview #DebuggingTips
Видео BST with Dead End | GfG POTD | 09-06-2025 | GfG Problem of the day | GeeksforGeeks канала LeetGeek
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