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:
- The range of node's value is in the range of 32-bit signed integer
/**
* 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;
double sum=0,count=0;
queue<TreeNode*>q;
q.push(root);
q.push(nullptr);
while(!q.empty())
{
TreeNode*t=q.front();
q.pop();
if(t==nullptr)
{
ret.push_back(sum/count);
sum=0;
count=0;
if(!q.empty())
q.push(nullptr);
}
else
{
sum+=t->val;
count++;
if(t->left)
q.push(t->left);
if(t->right)
q.push(t->right);
}
}
return ret;
}
};
//用nullptr分层,遇到nullptr表示一层结束,算出这一层的平均值添加到数组