## 题目描述

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.

## 解法一

### Python

``````# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
def averageOfLevels(self, root):
"""
:type root: TreeNode
:rtype: List[float]
"""
if root is None:
return []
queue = collections.deque()
result = []
queue.append(root)
while queue:
sum_current_level = 0
count_current_level = len(queue)
for _ in range(count_current_level):
current = queue.popleft()
sum_current_level += current.val
if current.left:
queue.append(current.left)
if current.right:
queue.append(current.right)
if count_current_level != 0:
result.append(sum_current_level / count_current_level)
return result``````

### Java

``````/**
* 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) {
if (root == null) {
return new ArrayList<>();
}
Queue<TreeNode> queue = new ArrayDeque<>();
List<Double> result = new ArrayList<>();
while (!queue.isEmpty()) {
long sumCurrentLevel = 0;
int countCurrentLevel = queue.size();
for (int i = 0; i < countCurrentLevel; i++) {
TreeNode current = queue.poll();
sumCurrentLevel += current.val;
if (current.left != null) {
}
if (current.right != null) {
}
}
if (countCurrentLevel != 0) {
}
}
return result;
}
}``````

### C++

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<double> averageOfLevels(TreeNode *root) {
if (root == nullptr) {
return vector<double>();
}
queue<TreeNode *> que;
vector<double> result;
que.push(root);
while (!que.empty()) {
long sumCurrentLevel = 0;
int countCurrentLevel = que.size();
for (int i = 0; i < countCurrentLevel; i++) {
TreeNode *current = que.front();
que.pop();
sumCurrentLevel += current->val;
if (current->left != nullptr) {
que.push(current->left);
}
if (current->right != nullptr) {
que.push(current->right);
}
}
if (countCurrentLevel != 0) {
result.push_back((double) sumCurrentLevel / countCurrentLevel);
}
}
return result;
}
};``````