Hexo

Question

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<sup>-5</sup> of the actual answer will be accepted.

Solution

BFS搜索，记录单层的总和sum和单层的个数count。
每次遍历一个层级的所有节点，并更新sum和count。
遍历完毕后将当层级的平均数加入列表，同时将sum和count清零。

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        
        double sum = 0, count = 0;
        int level = 1;
        q.add(root);
        
        while(!q.isEmpty()){
    
            for(int i = 0; i < level; i++){
                TreeNode curr = q.poll();
                if(curr.left != null) q.add(curr.left);
                if(curr.right != null) q.add(curr.right);
                count++;
                sum += curr.val;
            }
            res.add(sum/count);
            sum = 0;
            count = 0;
            level = q.size();
        }
        return res;
    }
}