1.题目描述：

给定一个非空二叉树的根节点root，以数组的形式返回每一层节点的平均值。与实际答案相差10-5以内的答案可以被接受。

2.BFS层序遍历：

看完题目毫不犹豫选择广度层序遍历，代码很容易写。

/*** 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) {if (root == null) return null;List<Double> list = new ArrayList<>();Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);while (!queue.isEmpty()) {int size = queue.size();double sum = 0;for (int i = 0; i < size; i++) {TreeNode temp = queue.poll();sum += temp.val;if (temp.left != null) queue.add(temp.left);if (temp.right != null) queue.add(temp.right);}list.add(sum / size);           }return list;}
}

3.DFS深度优先搜索遍历：

class Solution {public List<Double> averageOfLevels(TreeNode root) {List<Integer> counts = new ArrayList<Integer>();//用来维护每一层节点的数量List<Double> sums = new ArrayList<Double>();//用来维护每一层节点的平均值dfs(root, 0, counts, sums);List<Double> averages = new ArrayList<Double>();int size = sums.size();for (int i = 0; i < size; i++) averages.add(sums.get(i) / counts.get(i));return averages;}public void dfs(TreeNode root, int level, List<Integer> counts, List<Double> sums) {if (root == null) return;if (level < sums.size()) {sums.set(level, sums.get(level) + root.val);//更新节点的平均值counts.set(level, counts.get(level) + 1);//更新节点的数量} else {sums.add(1.0 * root.val);counts.add(1);}dfs(root.left, level + 1, counts, sums);dfs(root.right, level + 1, counts, sums);}
}