Maximum width of binary tree

Time: O(N); Space: O(H); med

Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a full binary tree, but some nodes are null.

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.

Example 1:

Input: root = {TreeNode} [1,3,2,5,3,None,9]

    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: root = {TreeNode} [1,3,None,5,3]

    1
   /
  3
 / \
5   3

Output: 2

Explanation:

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

Example 3:

Input: root = {TreeNode} [1,3,2,5]

    1
   / \
  3   2
 /
5

Output: 2

Explanation:

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

Example 4:

Input: root = {TreeNode} [1,3,2,5,None,None,9,6.None,None,None,None,None,None,7]

      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).

Note:

  • Answer will in the range of 32-bit signed integer.

[1]:

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

[2]:

class Solution1(object):
    def widthOfBinaryTree(self, root) -> int:
        """
        :type root: TreeNode
        :rtype: int
        """
        def dfs(node, i, depth, leftmosts):
            if not node:
                return 0
            if depth >= len(leftmosts):
                leftmosts.append(i)
            return max(i-leftmosts[depth]+1, \
                       dfs(node.left, i*2, depth+1, leftmosts), \
                       dfs(node.right, i*2+1, depth+1, leftmosts))

        leftmosts = []
        return dfs(root, 1, 0, leftmosts)

[3]:

s = Solution1()

root = TreeNode(1)
root.left, root.right = TreeNode(3), TreeNode(2)
root.left.left, root.left.right = TreeNode(5), TreeNode(3)
root.right.right = TreeNode(9)
assert s.widthOfBinaryTree(root) == 4

root = TreeNode(1)
root.left = TreeNode(3)
root.left.left, root.left.right = TreeNode(5), TreeNode(3)
assert s.widthOfBinaryTree(root) == 2

root = TreeNode(1)
root.left, root.right = TreeNode(3), TreeNode(2)
root.left.left = TreeNode(5)
assert s.widthOfBinaryTree(root) == 2

root = TreeNode(1)
root.left, root.right = TreeNode(3), TreeNode(2)
root.left.left = TreeNode(5)
root.left.left.left = TreeNode(6)
root.right.right = TreeNode(9)
root.right.right.right = TreeNode(7)
assert s.widthOfBinaryTree(root) == 8