Maximum Width of Binary Tree

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:
# 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 inorder(self, root, l, w):
        if root:
            self.inorder(root.left, l + 1, 2 * w)
            if l in self.levels:
                self.levels[l] = (min(self.levels[l][0], w), max(self.levels[l][1], w))
                self.mwidth = max(self.mwidth, self.levels[l][1] - self.levels[l][0] + 1)
            else:
                self.levels[l] = (w, w)
                self.mwidth = max(self.mwidth, 1)
            self.inorder(root.right, l + 1, 2 * w + 1)
    
    def widthOfBinaryTree(self, root: Optional[TreeNode]) -> int:
        self.levels = {}
        self.mwidth = 0
        self.inorder(root, 0, 0)
        return self.mwidth

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