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.

思路：首先要使用层次遍历，因为每次遍历后要计算对应层的平均值。所以又需要加上对层次的标记。

/**
 * 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) {
        //存放平均值的list
        List<Double> list = new ArrayList<>();
        //存放树的队列，用于层次遍历。
        Queue<TreeNode> queue = new ArrayDeque<>();
        // 存放对应树的层次信息，便于计算均值。
        Queue<Integer> queue1 = new ArrayDeque<>();
        // 初始层级h=0
        int h = 0;
        // 对应层级的节点数
        int n = 0;
        // 对应层级的节点数之和，要用double型。
        double sum = 0;

        queue.offer(root);
        // 初始层级放入后+1，根节点只有一个。
        queue1.offer(new Integer(h++));
        while(!queue.isEmpty()){
            root = queue.poll();
            int hh = queue1.poll().intValue();
            // 如果处于同一层，需要将层级+1，并完成平均数计算，放入list
            if(h == hh){
                h++;
                list.add(new Double(sum/n));
                sum = 0;
                n = 0;
            }
            sum += root.val;
            n++;
            // 层次遍历，放入左右子树和对应层级
            if(root.left != null){
                queue.offer(root.left);
                queue1.offer(new Integer(h));
            }
            if(root.right != null){
                queue.offer(root.right);
                queue1.offer(new Integer(h));
            }
        }
        // 最后一次的均值未计算，要单独处理
        list.add(new Double(sum/n));
        return list;
    }
}