662. Maximum Width of Binary Tree

https://leetcode.com/problems/maximum-width-of-binary-tree/

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: 

           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).

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

Thoughts

问二叉树最大的宽度，看作full tree后每层最外结点位置相减。按层=>BFS。full binary tree上左右子结点所在位置为父节点位置*2 + 1和+2，再每层最右减去最左的位置。

Code

# 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:
        if not root: return 0
        res, q = 0, collections.deque()
        q.append((root, 0))
        while len(q) > 0:
            _, l = q[0]
            for i in range(len(q)):
                t, p = q.popleft()
                if t.left: q.append((t.left, 2 * p + 1))
                if t.right: q.append((t.right, 2 * p + 2))
            res = max(res, p - l + 1)
        return res
        

Analysis

时间复杂度为一次遍历，O(n)