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   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.

解题思路：

使用queue，遍历树的每一层，确保每次queue中只保存了一层的节点。

代码：

/**
 * 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> ret;
        queue<TreeNode*> nodes;
        nodes.push(root);
        while (!nodes.empty()) {
            double sum = 0;
            int size = nodes.size();
            for (int i = 0; i < size; ++i) {
                TreeNode* tmp = nodes.front();
                nodes.pop();
                sum += tmp->val;
                if (tmp->left)
                    nodes.push(tmp->left);
                if (tmp->right)
                    nodes.push(tmp->right);
            }
            ret.push_back(sum / size);
        }
        return ret;
    }
};