UNC: Data Structures - S24 - Lecture 10 - Doubly LL, Binary Search Trees, complexity of operations

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🎯 Key Takeaways for quick navigation:

00:14 *🌐 The course progresses into real data structures after the first midterm, starting with the exploration of doubly linked lists.*
01:08 *🌲 Trees, being two-dimensional and non-linear data structures, are introduced, and their complexities will be analyzed throughout the course.*
04:30 *🔄 Comparison of memory size emphasizes the significance of logarithmic complexity (log n) over linear complexity (n) for search operations in large data structures.*
07:02 *🧠 The lecture highlights the importance of selecting the right data structure for efficient search engine operations, exemplifying Google's search speed with its massive database.*
09:26 *🔄 Introduction of doubly linked lists (W link lists) as a stepping stone towards more complex data structures like trees.*
13:10 *📉 Doubly linked lists offer advantages in deletion operations but come with the disadvantage of increased space complexity (50% more space per node).*
15:09 *🌳 The concept of trees is introduced, emphasizing the downward growth with a root node and leaves, and defining characteristics such as siblings and parent-child relationships.*
18:06 *🧑‍👦‍👦 Discussing the organization of a tree structure, considering the inclusion of a parent pointer, a root pointer, and the identification of leaves.*
21:36 *🌐 The challenge of an arbitrary number of children is introduced, leading to the consideration of a "number of children" field in each node, prompting further thoughts on how to organize subsequent pointers.*
23:01 *🌳 A binary tree allows either 0, 1, or 2 children per node, simplifying the structure compared to multiple children.*
25:56 *🔄 Binary trees come with a 300% overhead in terms of memory due to pointers, which can be a disadvantage.*
32:35 *🔍 Binary search trees (BST) leverage the structure by organizing data such that left child nodes contain smaller values, and right child nodes contain larger values.*
39:16 *🌲 A perfect binary tree has all leaf nodes at the same level and each non-leaf node has exactly two children.*
41:07 *🔄 The order of insertion in a binary search tree influences whether it becomes a perfect binary tree or an unbalanced tree.*
43:47 *⚖️ Balanced trees aim to maintain similar heights for left and right subtrees, simplifying complexity operations like search, insert, and delete.*
46:18 *🧠 Understanding the intuition in data structures is crucial for retention and problem-solving; intuition aids in grasping concepts more effectively.*
48:23 *🌲 In a balanced binary search tree, the search complexity is O(log n) due to the recursive nature of narrowing down the search space by half.*
52:06 *📊 The search complexity in terms of height (H) is O(H), which, when balanced, simplifies to O(log n), indicating a significant reduction compared to a linear search.*
57:25 *📈 Establishing a relationship between the height (H) and the number of nodes (N) in a perfect binary tree leads to H = log₂(N + 1) - 1, demonstrating a logarithmic relationship.*
58:35 *🔄 The search operation's complexity is O(log n) in a balanced binary search tree, showcasing efficiency compared to linear search and even modified linked lists.*
01:00:55 *➕ Insertion in a balanced binary search tree maintains an O(log n) complexity, similar to the search operation, as it involves traversing the height of the tree.*
01:07:42 *➖ Deletion in a balanced binary search tree has a complexity of O(log n) or O(H), encompassing the search and the deletion process, which is proportional to the height of the tree.*
01:10:52 *🔄 Deleting a node with a single child involves updating the parent's pointer to the child, resulting in O(log n) complexity.*
01:11:06 *🌲 Deleting a node with two children, like deleting node 77, requires finding the smallest element in the right sub-tree (92) or the largest in the left sub-tree (72) to replace the target node, ensuring balance.*
01:12:32 *🔄 When replacing a node with two children, the replacement value (e.g., 92) is determined by being the smallest in the right sub-tree or the largest in the left sub-tree.*