637. Average of Levels in Binary Tree

Q:

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.

A:

借用了两个数组，记录sum , and count;  

/**
 * 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> res;
        vector<int> count;
        helper(root, res, count, 1);
        for(int i =0;i<count.size(); i++)
        {
            res[i] /= count[i];
        }
        return res;
    }
private:
    void helper(TreeNode* root, vector<double> & res, vector<int> &count, int depth)
    {
        if(!root)
            return;
        if(res.size() < depth)
        {
            count.push_back(0);
            res.push_back(0);  // at most need one layer
        }
        res[depth-1] += root->val;
        count[depth-1] ++;
        helper(root->left, res, count, depth+1);
        helper(root->right, res,count, depth+1);
    }
};

借用了两个数组，记录sum , and count;