The problem statement, as given on LeetCode, asks us to find the maximum width of a binary tree. The width of a level is defined as the number of nodes at that level. The maximum width of the binary tree is the maximum width among all levels.

Initialization: Start with the root node of the binary tree. Initialize a queue to hold nodes and their positions. Set a variable max_width to keep track of the maximum width seen so far (initially set to 0).

Breadth-First Search (BFS): Perform a BFS traversal of the binary tree. Begin with the root node and assign it a position of 0. Enqueue the root node along with its position into the queue.

While the Queue is not Empty:

Get the size of the queue, which represents the number of nodes at the current level.

Initialize left_index and right_index to track the indices of the leftmost and rightmost nodes at the current level.

Iterate through the nodes at the current level:

Dequeue a node and its position from the queue.

Update left_index with the position of the leftmost node.

Update right_index with the position of the last node (rightmost) in the current level.

Calculate the width of the current level as right_index - left_index + 1.

Update max_width with the maximum of its current value and the width of the current level.

If the node has a left child, enqueue the left child with a position calculated as (2 * current_position) + 1.

If the node has a right child, enqueue the right child with a position calculated as (2 * current_position) + 2.

Return max_width: After processing all levels, max_width will hold the maximum width of the binary tree.

Time Complexity

The time complexity is O(N), where N is the number of nodes in the binary tree. This is because both solutions traverse all nodes in the tree once using BFS.