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.
/**
* 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) {
var res = new List<double>();
var queue = new Queue<TreeNode>();
queue.Enqueue(root);
while(queue.Count()>0){
int count = queue.Count();
int i = count;
double sum = 0;
while(i-->0){
var node = queue.Dequeue();
sum += node.val;
if(node.left != null){
queue.Enqueue(node.left);
}
if(node.right != null){
queue.Enqueue(node.right);
}
}
res.Add(sum/count);
}
return res;
}
}Time Complexity: O(n)
Space Complexity: O(n)
