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662. Maximum Width of Binary Tree

Difficulty Topics

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:

Input: root = [1,3,2,5,3,null,9]
Output: 4
Explanation: The maximum width exists in the third level with length 4 (5,3,null,9).

Example 2:

Input: root = [1,3,2,5,null,null,9,6,null,7]
Output: 7
Explanation: The maximum width exists in the fourth level with length 7 (6,null,null,null,null,null,7).

Example 3:

Input: root = [1,3,2,5]
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

Solution

maximum-width-of-binary-tree.py
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def widthOfBinaryTree(self, root: Optional[TreeNode]) -> int:
        queue = deque([root])
        mp = defaultdict(int)
        mp[root] = 1
        res = 1

        while queue:
            n = len(queue)
            start = end = -1

            for index in range(n):
                node = queue.popleft()

                if index == 0:
                    start = mp[node]
                elif index == n - 1:
                    end = mp[node]

                if node.left:
                    queue.append(node.left)
                    mp[node.left] = 2 * mp[node]

                if node.right:
                    queue.append(node.right)
                    mp[node.right] = 2 * mp[node] + 1

            res = max(res, end - start + 1)

        return res