Average of Levels in Binary Tree问题及解法

问题描述：

Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

示例：

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].

问题分析：

根据BFS遍历，求得每一层的的平均值。

过程详见代码：

/**
* 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) {
queue<TreeNode*> q;
vector<double> res;
q.push(root);
while (!q.empty())
{
double sum = 0.0;
int count = q.size();
queue<TreeNode*> tq;
while (!q.empty())
{
TreeNode* node = q.front();
q.pop();
sum += node->val;
if (node->left != NULL)
{
tq.push(node->left);
}
if (node->right != NULL)
{
tq.push(node->right);
}
}
res.push_back(sum / count);
q = tq;
}
return res;
}
};