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.
Related Topics:
Tree
Similar Questions:
// OJ: https://leetcode.com/problems/average-of-levels-in-binary-tree/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<double> averageOfLevels(TreeNode* root) {
if (!root) return {};
vector<double> ans;
queue<TreeNode*> q;
q.push(root);
while (q.size()) {
double cnt = q.size(), sum = 0;
for (int i = 0; i < cnt; ++i) {
root = q.front();
q.pop();
sum += root->val;
if (root->left) q.push(root->left);
if (root->right) q.push(root->right);
}
ans.push_back(sum / cnt);
}
return ans;
}
};