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 { * 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<>(); q.add(root); while(!q.isEmpty()) { int n=q.size();//很巧妙的计算每层个数，只有使用完该层才会进入到下一层实验 double sum=0.0; for(int i=0;i<n;i++) { TreeNode x=q.poll(); sum+=x.val; if(x.left!=null)q.add(x.left); if(x.right!=null)q.add(x.right); } result.add(sum/n); } return result; } } 一个简单的广度优先搜索，难点在于思考广度优先搜索每一层的个数鉴定
