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    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:

    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.
    Solution
    Approach 1: Using Classical BFS
    Here we used Queue with Poll and Offer function for managing the queue.
    /**
     * 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> result = new ArrayList<>();
    Queue<TreeNode> q = new LinkedList<>();
    
    if(root==null)return result;
    
    q.offer(root);
    
    while (!q.isEmpty()){
              double sum=0;
     int n= q.Size();
    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;
    
    }

     

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