How do you solve Maximum Width of Binary Tree on LeetCode?

Maximum Width of Binary Tree (LeetCode #662) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The width of a level can be calculated using indices in a complete binary tree format.

What is the brute force solution for Maximum Width of Binary Tree?

The brute force approach for Maximum Width of Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates the width of each level by traversing the entire tree level by level, counting the non-null nodes and considering the gaps. This method is straightforward but inefficient for larger trees. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Width of Binary Tree?

Maximum Width of Binary Tree uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Binary Tree. The recommended patterns to study are: Breadth-First Search, Level Order Traversal.

What are the interview tips for Maximum Width of Binary Tree?

Clarify the definition of width with the interviewer. Discuss the approach before coding to ensure understanding. Consider edge cases, such as trees with only one node or very unbalanced trees.

What are common mistakes when solving Maximum Width of Binary Tree?

Not considering null nodes in the width calculation. Failing to maintain the correct indices during traversal.

#662

Maximum Width of Binary Tree

Medium

TreeDepth-First SearchBreadth-First SearchBinary TreeBreadth-First SearchLevel Order Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses a level-order traversal with indices to calculate the width efficiently. By treating the indices as positions in a complete binary tree, we can compute the width without needing to traverse all nodes at each level.

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Algorithm

3 steps

1Step 1: Use a queue to perform a level-order traversal of the tree, storing each node along with its index.
2Step 2: For each level, calculate the width using the first and last indices of the non-null nodes.
3Step 3: Update the maximum width found during the traversal.

solution.py18 lines

1# Full working Python code
2from collections import deque
3
4def widthOfBinaryTree(root):
5    if not root:
6        return 0
7    max_width = 0
8    queue = deque([(root, 0)])
9    while queue:
10        level_length = len(queue)
11        _, first_index = queue[0]
12        for _ in range(level_length):
13            node, index = queue.popleft()
14            if node:
15                queue.append((node.left, 2 * index))
16                queue.append((node.right, 2 * index + 1))
17        max_width = max(max_width, index - first_index + 1)
18    return max_width

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the queue storing nodes at each level.

1The width of a level can be calculated using indices in a complete binary tree format.
2Using a queue allows for efficient level-order traversal.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.