637. Average of Levels in Binary Tree

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Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

Example 1:

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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.

解法1：

一个bfs解决问题。
要注意的是当中的加和可能会overflow，要用long。
C++

1

Java

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<Double>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            long sum = 0;
            for (int i = 0; i < size; i++) {
                TreeNode current = queue.poll();
                sum += current.val;
                if (current.left != null) {
                    queue.offer(current.left);
                }
                if (current.right != null) {
                    queue.offer(current.right);
                }
            }
            res.add((double)sum / size);
        }
        return res;
    }
}