Maximum Width of Binary Tree
Problem Description
Given the root of a binary tree, return the maximum width of the given tree.
The maximum width of a tree is the maximum width among all levels.
The width of one level is defined as the length between the end-nodes (the leftmost and rightmost non-null nodes), where the null nodes between the end-nodes that would be present in a complete binary tree extending down to that level are also counted into the length calculation.
It is guaranteed that the answer will in the range of a 32-bit signed integer.
Example 1:
Output: 4
Explanation: The maximum width exists in the third level with length 4 (5,3,null,9).
Example 2:
Output: 7
Explanation: The maximum width exists in the fourth level with length 7 (6,null,null,null,null,null,7).
Example 3:
Output: 2
Explanation: The maximum width exists in the second level with length 2 (3,2).
Constraints:
- The number of nodes in the tree is in the range
[1, 3000]. -100 <= Node.val <= 100
Solve it on LeetCode
Approaches
1. BFS with Index Tracking
Intuition:
To determine the maximum width of the binary tree, one intuitive method is to perform a level-wise traversal while keeping track of the indices of nodes if they were part of a "complete" binary tree. The idea here is that if two nodes are on the same level, the difference between their indices can give us the width at that level. The maximum width found across all levels is the answer.
Steps:
- Use a breadth-first search (BFS) approach to visit all nodes level by level, using a queue to store each node along with its index as if it were in a complete binary tree.
- For each level:
- Compute the width as the difference between the index of the first and last nodes at that level plus one.
- Keep track of the maximum width encountered.
- Continue until all levels have been processed.
Code:
- Time Complexity: O(N), where N is the number of nodes since each node is processed exactly once.
- Space Complexity: O(N), due to the queue storing at most the maximum number of nodes in a given level.