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.

M1: BFS

注意！sum要用long，用int會溢位

time: O(n), space: O(N)　　-- 最多一層節點的個數

/**
 * 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<>();
        if(root == null) {
            return res;
        }
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while(!q.isEmpty()) {
            
 
long sum = 0;
            int size = q.size();
            for(int i = 0; i < size; i++) {
                TreeNode t = q.poll();
                sum += t.val;
                if(t.left != null) {
                    q.offer(t.left);
                }
                if(t.right != null) {
                    q.offer(t.right);
                }
            }
            res.add((double)sum / size);
        }
        return res;
    }
}

M2: DFS

time: O(n), space: O(height)