LeetCode 637. Average of Levels in Binary Tree(二叉树每层的平均值)
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:
- The range of node's value is in the range of 32-bit signed integer.
解法一
思路
这题是典型的BFS了,只需要在遍历当前层级的时候记录下结点个数和值的累加和就好了。
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<>();
queue.add(root);
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) {
queue.add(current.left);
}
if (current.right != null) {
queue.add(current.right);
}
}
if (countCurrentLevel != 0) {
result.add((double) sumCurrentLevel / countCurrentLevel);
}
}
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;
}
};