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.
Solution:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Double> averageOfLevels(TreeNode root) {
Queue<TreeNode> queue = new ArrayDeque();
queue.offer(root);
List<Double> result = new ArrayList();
while (!queue.isEmpty()) {
int size = queue.size();
long sum = 0;
for (int i = 0; i < size; i ++) {
TreeNode curr = queue.poll();
sum += curr.val;
if (curr.left != null) queue.offer(curr.left);
if (curr.right != null) queue.offer(curr.right);
}
result.add((double) sum / size);
}
return result;
}
}