题目描述：
Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.
Example 1:
Input:
3
/ \
9 20
/ \
15 7
Output: [3, 14.5, 11]
Explanation:
The average value of nodes on level 0 is 3, on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].
Note:
The range of node's value is in the range of 32-bit signed integer.
做题思路：
利用层次遍历，重点在于分层，分出层之后就能求出每一层的平均值了。时间复杂度O（n）
答案详解：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<double> averageOfLevels(TreeNode* root) {
vector<double> a;
TreeNode* p = root;
int rear = -1;//生命队列的前后位置
int front = -1;
int last = 0;//当front等于last时，level++
int level = 0;
struct TreeNode* q[10001];//将节点存在队列里面
double ave;//每层的平均数
int i = 0;//每一层有几个节点
if(root=NULL){
a.push_back(NULL);
return a;
}
q[++rear] = p;//初始化第一个节点
i=0;
while(front<rear)
{
ave = ave + q[++front]->val;
cout<< ave << endl;
p = q[front];
i++;
if(p->left){
q[++rear] = p->left;
}
if(p->right){
q[++rear] = p->right;
}
if( front == last)
{
a.push_back(ave/i);
last = rear;
ave = 0;
i=0;
}
}
return a;
}
};
