Average of Levels in Binary Tree

Problem

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.

Solution

Typical BFS solution

public class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        if (root == null) return new ArrayList<>();
        List<Double> res = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        while (!q.isEmpty()) {
            int n = q.size();
            double sum = 0; //We need to add double in our result list 
            for (int i = 0; i < n; i++) {
                TreeNode node = q.poll();
                sum += node.val;
                if(node.left != null) q.push(node.left);
                if (node.right != null) q.push(node.right);
            }
            res.add(sum / n);
        }
        return res;
    }
}

Analysis

This is also a typical traversal problem
We need to go through each level to get the sum values of each node
Therefore, we use BFS with a queue to accomplish it
Notice that we need to check null before we pushing the left or right into our queue