Maximum Width of Binary Tree

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given a binary tree.
  • We need to find the maximum width of the tree, which is defined as the maximum number of nodes in any level of the tree.

2. Analyze the problem: To solve this problem, we can perform a level order traversal of the tree and keep track of the width at each level. We can use a queue data structure to traverse the tree level by level.

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Initialize a variable max_width to 0.
  2. Create an empty queue and enqueue the root of the binary tree.
  3. While the queue is not empty:
    • Get the current size of the queue to represent the number of nodes at the current level.
    • Update max_width if the current level width is greater than max_width.
    • Enqueue all the children of the nodes in the current level.
    • Dequeue the nodes in the current level.
  4. Return max_width.

4. Explain your approach: The approach involves performing a level order traversal of the binary tree using a queue. We keep track of the width at each level and update the max_width variable if the current level width is greater.

5. Write clean and readable code:

python
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def maxWidthOfBinaryTree(root):
    if not root:
        return 0

    max_width = 0
    queue = [(root, 0)]

    while queue:
        level_width = queue[-1][1] - queue[0][1] + 1
        max_width = max(max_width, level_width)

        next_level = []
        for node, pos in queue:
            if node.left:
                next_level.append((node.left, 2 * pos))
            if node.right:
                next_level.append((node.right, 2 * pos + 1))

        queue = next_level

    return max_width

6. Test your code: Let's test the code with some test cases:

  • Test case 1:

    • root = TreeNode(1) /
      3 2 / \ \
      5 3 9
    • The expected output is 4 because the maximum width is at the second level, which has 4 nodes.

  • Test case 2:

    • root = TreeNode(1) /
      3 2 / \
      5 9 / \ /
      6 7 8 10
    • The expected output is 8 because the maximum width is at the fourth level, which has 8 nodes.
python
# Test case 1
root1 = TreeNode(1)
root1.left = TreeNode(3)
root1.right = TreeNode(2)
root1.left.left = TreeNode(5)
root1.left.right = TreeNode(3)
root1.right.right = TreeNode(9)
print(maxWidthOfBinaryTree(root1))  # Expected output: 4

# Test case 2
root2 = TreeNode(1)
root2.left = TreeNode(3)
root2.right = TreeNode(2)
root2.left.left = TreeNode(5)
root2.right.right = TreeNode(9)
root2.left.left.left = TreeNode(6)
root2.left.left.right = TreeNode(7)
root2.right.right.left = TreeNode(8)
root2.right.right.right = TreeNode(10)
print(maxWidthOfBinaryTree(root2))  # Expected output: 8

7. Optimize if necessary: The current solution is already optimal with a time complexity of O(n), where n is the number of nodes in the binary tree. We traverse each node once using the queue. The space complexity is also O(n) as we store at most n/2 nodes in the queue when the tree is a complete binary tree.

8. Handle error cases: The code assumes that the input root is a valid binary tree, but if it is None, the code will return 0, indicating an empty tree.

9. Discuss complexity analysis: The time complexity of the solution is O(n), where n is the number of nodes in the binary tree. We traverse each node once using the queue. The space complexity is also O(n) as we store at most n/2 nodes in the queue when the tree is a complete binary tree. The solution has linear time and space complexity.

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