LeetCode 637. Average of Levels in Binary Tree二叉树的层平均值 (C++)

题目：

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:

1. The range of node's value is in the range of 32-bit signed integer.

分析：

题目很好理解，就是求二叉树每一层的平均值。

我们可以创建一个用来存储每层节点值和的数组，和一个存储每层节点个数的数组，通过递归的方式来求二叉树的层平均值，注意在求当前层时要判断层数是否超过了数组的大小，一旦超过，就要扩大数组，一遍存储当前层的信息。

程序：

/**
 * 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) {
        getAverage(root, res, );
        for(int i = ; i < res.size(); ++i){
            res[i] /= nums[i];
        }
        return res;
    }
    void getAverage(TreeNode* root, vector<double> &res, int level){
        if(root == nullptr) return;
        if(level >= res.size()){
            res.push_back();
            nums.push_back();
        }
        res[level] += root->val;
        nums[level]++;
        getAverage(root->left, res, level+);
        getAverage(root->right, res, level+);
    }
private:
    vector<double> res;
    vector<int> nums;
};