LeetCode 637. Average of Levels in Binary Tree(Easy)-优快云博客

本文介绍了一种算法，用于计算给定非空二叉树各层节点的平均值，并以数组形式返回。通过使用队列进行层次遍历，算法能够有效地处理32位整数范围内的值，避免了整数溢出的问题。

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:

The range of node’s value is in the range of 32-bit signed integer.

Analysis:
这题需要注意的是node中提到的整数范围，因为int能表示的最大整数是 $2^{31}-1$ ，因此这道题可能会出现多个接近 $2^{32}-1$ 的整数相加造成整数溢出的现象，简单的解决方法是将变量sum的类型设置为long。long能表示的最大整数是 $2^{63}-1$ .

Code:

/**
 * 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> avgs = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while(q.peek() != null) {
            int size = q.size();
            long sum = 0;
            for(int i = 0; i < size; i++) {
                TreeNode node = q.poll();
                sum += node.val;
                if(node.left != null) {
                    q.offer(node.left);
                }
                if(node.right != null) {
                    q.offer(node.right);
                }
            }
            double avg = sum * 1.0 / size;
            avgs.add(avg);
        }
        return avgs;
    } 
}