637. Average of Levels in Binary Tree

https://leetcode.com/problems/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.

X. BFS
https://leetcode.com/articles/average-of-levels/
https://discuss.leetcode.com/topic/95214/java-bfs-solution

Classic bfs problem. At each level, compute the average since you already know the size of the level.

public List<Double> averageOfLevels(TreeNode root) {
    List<Double> result = new ArrayList<>();
    Queue<TreeNode> q = new LinkedList<>();
    
    if(root == null) return result;
    q.add(root);
    while(!q.isEmpty()) {
        int n = q.size();
        double sum = 0.0;
        for(int i = 0; i < n; i++) {
            TreeNode node = q.poll();
            sum += node.val;
            if(node.left != null) q.offer(node.left);
            if(node.right != null) q.offer(node.right);
        }
        result.add(sum / n);
    }
    return result;
}

X. DFS
https://discuss.leetcode.com/topic/95567/java-solution-using-dfs-with-full-comments

    public List<Double> averageOfLevels(TreeNode root) {
        // list answer for sum all value in same level
        List<Double> answer = new ArrayList<Double>();
        
        // list counter for count number of node in same level
        List<Integer> counter = new ArrayList<Integer>();
        
        // using dfs to sum all value in same level and count number of node in same level
        dfs(0, root, answer, counter);
        
        // answer will be answer[level] / counter[level]
        for (int level = 0; level < answer.size(); level++) {
            answer.set(level, answer.get(level) / counter.get(level));
        }
        return answer;
    }

    public void dfs(int level, TreeNode node, List<Double> answer, List<Integer> counter) {
        if (node == null) {
            return;
        }

        if (answer.size() <= level) {
            answer.add(0.0);
            counter.add(0);
        }

        answer.set(level, answer.get(level) + node.val);
        counter.set(level, counter.get(level) + 1);

        // go left node and right node
        dfs(level + 1, node.left, answer, counter);
        dfs(level + 1, node.right, answer, counter);
    }