LeetCode 637. Average of Levels in Binary Tree

Problem

LeetCode 637. Average of Levels in Binary Tree

1. 题目简述

给出一棵树，每个节点都是signed int类型，计算出每一层的平均值（double类型），将其放入一个list中后返回。例如：

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].

2. 算法思路

Queue + BFS

首先，我们要想好用什么方式对树进行遍历，BFS or DFS。这里涉及到层数，一层一层来计算，因此BFS相对更好一些。DFS则需要用Queue来对节点进行存储，问题是怎么样对不同层级来计算，这里在循环每一层之前，用一个size变量来记录当前Queue大小，这个大小就是这一层层级的节点数。

3. 解法

1. 注意使用变量记录当前界定个数，方便循环。

/**
 * 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> result = new ArrayList<Double>();

        if (root == null) {
            return result;
        }

        Queue<TreeNode> queue1 = new LinkedList<TreeNode>();
        queue1.add(root);

        while (!queue1.isEmpty()) {
            int size = queue1.size();
            double sum = 0.0;

            for (int i = 0; i < size; i++) {
                TreeNode temp = queue1.poll();
                if (temp.left != null) {
                    queue1.offer(temp.left);
                }
                if (temp.right != null) {
                    queue1.offer(temp.right);
                }
                sum += temp.val;
            }

            result.add(new Double(sum / size));
        }

        return result;
    }
}