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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.

Solution (Runtime: 6 ms)


/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Double> averageOfLevels(TreeNode root) {
List<Double> rst = new ArrayList<>();


if(root == null) {
return rst;
}

Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);

while(!queue.isEmpty()) {
int size = queue.size();
double 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);
}
}

rst.add(sum / size);
}

return rst;
}
}

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