题目如下：
Given a binary tree, determine if it is a complete binary tree.
Definition of a complete binary tree from :
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.
Example 1:
Input: [1,2,3,4,5,6]Output: trueExplanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.Example 2:
Input: [1,2,3,4,5,null,7]Output: falseExplanation: The node with value 7 isn't as far left as possible.Note:
- The tree will have between 1 and 100 nodes.
解题思路：完全二叉树有这个一个定律：完全二叉树中任一节点编号n,则其左子树为2n,右子树为2n+1。所以，只要遍历一遍二叉树，记根节点的编号为1，依次记录每个遍历过的节点的编号，同时记录编号的最大值MAX。最后判断所有遍历过的节点的编号的和与sum(1~MAX)是否相等即可。
代码如下：
# Definition for a binary tree node.# class TreeNode(object):# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution(object): number = [] maxNum = 0 def traverse(self,node,seq): self.maxNum = max(self.maxNum,seq) self.number.append(seq) if node.left != None: self.traverse(node.left,seq*2) if node.right != None: self.traverse(node.right,seq*2+1) def isCompleteTree(self, root): """ :type root: TreeNode :rtype: bool """ self.number = [] self.maxNum = 0 self.traverse(root,1) return (self.maxNum+1)*self.maxNum/2 == sum(self.number)