637.AverageOfLevels.cs

// 637. Average of Levels in Binary Tree

// Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. Answers within 10-5 of the actual answer will be accepted.
 

// Example 1:


// Input: root = [3,9,20,null,null,15,7]
// Output: [3.00000,14.50000,11.00000]
// 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].
// Example 2:


// Input: root = [3,9,20,15,7]
// Output: [3.00000,14.50000,11.00000]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public IList<double> AverageOfLevels(TreeNode root) {
        IList<double> res = new List<double>();
        if (root == null)
        {
        return res;
        }

        Queue<TreeNode> queue = new Queue<TreeNode>();
        queue.Enqueue(root);
        while (queue.Count > 0) 
        {
            IList<int> level = new List<int>();
            int size = queue.Count;
            for (int i = 0; i < size; i++)
            {
                TreeNode currNode = queue.Dequeue();
                level.Add(currNode.val);
                if (currNode.left != null)
                    queue.Enqueue(currNode.left);
                if (currNode.right != null)
                    queue.Enqueue(currNode.right);
            }
            double average = level.Average();
            res.Add(average);
        }
        
        return res;
    }
}