662. Maximum Width of Binary Tree 二叉树最大宽度

题目地址：https://leetcode.com/problems/maximum-width-of-binary-tree/description/

题目描述：

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.

Thewidth 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.

题目大意

给定二叉树，求二叉树的最大宽度。二叉树某层的宽度是指其最左非空节点与最右非空节点之间的跨度。

解题方法

做法是层次遍历 + 完全二叉树的节点位置性质。这个性质指的是，每层都有 2 ^ (n-1)个节点。某节点的左孩子的标号是2n, 右节点的标号是2n + 1。因为这个题，中间缺少了节点的话，仍然要“认为”节点存在，所以需要使用这种标号的方法强制计算，而不是直接遍历。

遍历的方式是使用队列，其实很简单了。

代码：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def widthOfBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        queue = collections.deque()
        queue.append((root, 1))
        res = 0
        while queue:
            width = queue[-1][1] - queue[0][1] + 1
            res = max(width, res)
            for _ in range(len(queue)):
                n, c = queue.popleft()
                if n.left: queue.append((n.left, c * 2))
                if n.right: queue.append((n.right, c * 2 + 1))
        return res

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