Solution.py

"""

Given a binary tree, write a function to get 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 right most non-null nodes in the level, where the null nodes between the end-nodes are also counted into the length calculation.
It is guaranteed that the answer will in the range of 32-bit signed integer.
Example 1:

Input: 

           1
         /   \
        3     2
       / \     \  
      5   3     9 

Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).

Example 2:

Input: 

          1
         /  
        3    
       / \       
      5   3     

Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).

Example 3:

Input: 

          1
         / \
        3   2 
       /        
      5      

Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).

Example 4:

Input: 

          1
         / \
        3   2
       /     \  
      5       9 
     /         \
    6           7
Output: 8
Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).

 
Constraints:

The given binary tree will have between 1 and 3000 nodes.


"""


# 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: TreeNode) -> int:
        def fill_levels(node, levels, ci, cl):
            if node is None:
                return
            if len(levels) == cl:
                levels.append([ci, ci])
            else:
                levels[cl][0] = min(ci, levels[cl][0])
                levels[cl][1] = max(ci, levels[cl][1])
            fill_levels(node.left, levels, ci * 2 - 1, cl + 1)
            fill_levels(node.right, levels, ci * 2, cl + 1)
            
            
        levels = []
        fill_levels(root, levels, 1, 0)
        return max([l[1] - l[0] + 1 for l in levels])