Description:
Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.
Example:
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.
题目描述:
给定一个非空二叉树, 返回一个由每层节点平均值组成的数组.
示例 :
示例 1:
输入:
3
/ \
9 20
/ \
15 7
输出: [3, 14.5, 11]
解释:
第0层的平均值是 3, 第1层是 14.5, 第2层是 11. 因此返回 [3, 14.5, 11].
注意：
节点值的范围在32位有符号整数范围内。
思路:
参考LeetCode #107 Binary Tree Level Order Traversal II 二叉树的层次遍历 II
将每一层的结果记录求平均值即可
时间复杂度O(n), 空间复杂度O(n)
代码:
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 averageOfLevels(TreeNode* root) {
vector result;
queue q;
q.push(root);
while (q.size()) {
double total = 0;
int nums = q.size();
for (int i = 0; i < nums; i++) {
TreeNode* cur = q.front();
q.pop();
total += cur -> val;
if (cur -> left) q.push(cur -> left);
if (cur -> right) q.push(cur -> right);
}
result.push_back(total / nums);
}
return result;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List averageOfLevels(TreeNode root) {
List result = new ArrayList<>();
Queue q = new LinkedList<>();
q.offer(root);
while (!q.isEmpty()) {
double total = 0;
int nums = q.size();
for (int i = 0; i < nums; i++) {
TreeNode cur = q.poll();
total += cur.val;
if (cur.left != null) q.offer(cur.left);
if (cur.right != null) q.offer(cur.right);
}
result.add(total / nums);
}
return result;
}
}
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: TreeNode) -> List[float]:
queue, result = [root], []
while queue:
size, total = len(queue), 0
for _ in range(size):
item = queue.pop(0)
total += item.val
if item.left:
queue.append(item.left)
if item.right:
queue.append(item.right)
result.append(total / size)
return result