637. Average of Levels in Binary Tree

637. Average of Levels in Binary Tree

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.

题目大意：就是二叉树的同层级的结点的值的和求平均值

方法：一开始一直想着中序，左序，右序遍历，就想着拷贝一个相同的二叉树，想想很麻烦，就觉得应该是方法错误了。看了discuss的提示才想到BFS和DFS，，，看了二叉树都忘了、、这题BFS，存在队列里，就可以了

/**
 * 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 < > ();
        Queue < TreeNode > queue = new LinkedList < > ();
        queue.add(root);
        while (!queue.isEmpty()) {
            long sum = 0, count = 0;
            Queue < TreeNode > temp = new LinkedList < > ();
            while (!queue.isEmpty()) {
                TreeNode n = queue.remove();
                sum += n.val;
                count++;
                if (n.left != null)
                    temp.add(n.left);
                if (n.right != null)
                    temp.add(n.right);
            }
            queue = temp;
            res.add((double)sum/ count);
        }
        return res;
    }
}