637. 二叉树的层平均值

题目：

思路：

【1】这道题需要知道层次遍历这个理念，因为层次遍历就是把树的节点按照层级来输出，那么既然一层的数据都有了，对这一层进行平均数的统计也就简单了。

代码展示：

//时间4 ms 击败 8.92%
//内存43.4 MB 击败40.98%
//基于层次遍历来进行变动
/**
 * 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) {
        List<Double> res = new ArrayList<Double>();
        if (root == null) return res;

        Queue<TreeNode> que = new LinkedList<>();
        que.add(root);
        int count = que.size();
        TreeNode temp;
        while (!que.isEmpty()){
            List<Integer> resList = new ArrayList<>();
            while (count > 0){
                temp = que.poll();
                resList.add(temp.val);
                if (temp.left != null) que.add(temp.left);
                if (temp.right != null) que.add(temp.right);
                count--;
            }
            double sum = 0;
            for (int tem : resList){
                sum += tem;
            }
            res.add(sum/resList.size());
            count = que.size();
        }

        return res;
    }
}


//再简化一下一些不必要的操作
//时间2 ms 击败 92.18%
//内存43.5 MB 击败 22.17%
/**
 * 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) {
       List<Double> res = new ArrayList<Double>();
        if (root == null) return res;

        Queue<TreeNode> que = new LinkedList<>();
        que.add(root);
        int count = que.size();
        TreeNode temp;
        while (!que.isEmpty()){
            int size = que.size();
            double sum = 0;
            while (count > 0){
                temp = que.poll();
                sum += temp.val;
                if (temp.left != null) que.add(temp.left);
                if (temp.right != null) que.add(temp.right);
                count--;
            }
            res.add(sum/size);
            count = que.size();
        }

        return res;
    }
}