LeetCode 109 Conquered! Sorted List to Balanced BST (C++) - Efficient Conversion

Bridge the Gap: Transforming Sorted Linked Lists into Balanced BSTs in C++ (LeetCode 109)
Welcome, coders, to this in-depth guide on tackling LeetCode problem 109! We'll embark on a journey to convert a sorted linked list into a balanced Binary Search Tree (BST) in C++. This video equips you with the skills to bridge the gap between linear and hierarchical data structures, ensuring efficient searching and sorting operations.
Understanding the Problem:
LeetCode problem 109 presents an interesting challenge. You're given a singly linked list where each node holds an integer value, and the list is sorted in ascending order. The task is to convert this linked list into a balanced BST. Recall that a BST is a tree-based data structure where each node has a value, and its left subtree contains values less than the node's value, while the right subtree contains values greater than the node's value. Balancing ensures the tree's height is roughly logarithmic in the number of nodes, enabling efficient search and insertion operations.
Linked Lists vs. Binary Search Trees (BSTs):
Before diving into the solution, let's revisit the key differences between linked lists and BSTs:
Linked Lists: These linear data structures consist of nodes connected by links (pointers). Each node holds data (typically a value) and a reference to the next node in the sequence. Accessing elements by index is inefficient, but insertion and deletion at specific positions can be faster compared to BSTs.
BSTs: As mentioned earlier, BSTs are hierarchical structures where each node holds a value and pointers to its left and right child nodes. This structure allows for efficient searching by value, as you can quickly compare the search value with the current node's value to determine which subtree to explore next.
Visualizing Sorted List to Balanced BST:
Imagine a sorted linked list. We can convert this list into a balanced BST by finding the middle element and using it as the root node. The elements before the middle node become the left subtree, and the elements after the middle node become the right subtree. This process is applied recursively to both subtrees until the entire linked list is converted. We'll use clear diagrams throughout the video to illustrate this conversion.
Crafting the C++ Solution:
Now, let's explore how to solve the "Sorted List to Balanced BST" problem in C++. We'll present an efficient solution that leverages the sorted nature of the linked list and a fast and slow pointer approach to find the middle element.
Building the C++ Solution:
Defining ListNode and TreeNode structures for linked list nodes (with val and next pointer) and BST nodes (with val, left, and right pointers), respectively.
Creating a Solution class with a sortedListToBST method to convert the sorted linked list to a BST.
The sortedListToBST function takes the head of the linked list as input.
If the list is empty (head is null), return null (no BST to create).
If there's only one element (head.next is null), create a new BST node with that element's value and return it (becomes the root).
Call the helper function findMid to find the middle node of the linked list.
Create a new BST node with the middle node's value.
Recursively call sortedListToBST on the first half of the linked list (excluding the middle node) to construct the left subtree.
Recursively call sortedListToBST on the second half of the linked list (starting from the node after the middle node) to construct the right subtree.