637. Average of Levels in Binary Tree

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   7Output: [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:

1. 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;
    }
};