# 637. Average of Levels in Binary Tree

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.

``````class Solution {
public:
vector<double> averageOfLevels(TreeNode* root) {
if (!root) return {};
vector<double> res;
queue<TreeNode*> q{{root}};
while (!q.empty()) {
int n = q.size();
double sum = 0;
for (int i = 0; i < n; ++i) {
TreeNode *t = q.front(); q.pop();
sum += t->val;
if (t->left) q.push(t->left);
if (t->right) q.push(t->right);
}
res.push_back(sum / n);
}
return res;
}
};
``````

``````class Solution {
public:
vector<double> averageOfLevels(TreeNode* root) {
vector<double> res, cnt;
helper(root, 0, cnt, res);
for (int i = 0; i < res.size(); ++i) {
res[i] /= cnt[i];
}
return res;
}
void helper(TreeNode* node, int level, vector<double>& cnt, vector<double>& res) {
if (!node) return;
if (res.size() <= level) {
res.push_back(0);
cnt.push_back(0);
}
res[level] += node->val;
++cnt[level];
helper(node->left, level + 1, cnt, res);
helper(node->right, level + 1, cnt, res);
}
};
``````

Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal