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.

这个题和level order traversal完全一样，只是多了一步取平均值，还有一个要注意的事情是，某一层的数字加起来可能溢出，所以记得转成double再求和

/**
 * 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> res = new ArrayList<Double>();
        if(root == null)
            return res;

        LinkedList<TreeNode> curr = new LinkedList<TreeNode>();
        LinkedList<TreeNode> next = new LinkedList<TreeNode>();
        double sum = 0;
        int count = 0;

        curr.offer(root);

        while(curr.size() > 0) {
            TreeNode temp = curr.poll();
            sum += (double) temp.val;
            count++;

            if(temp.left != null)
                next.offer(temp.left);
            if(temp.right != null)
                next.offer(temp.right);

            if(curr.size() == 0) {
                res.add(sum / count);
                sum = 0.0;
                count = 0;
                curr = next;
                next = new LinkedList<TreeNode>();
            }
        }

        return res;
    }
}