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 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] 。
提示:
- 节点值的范围在32位有符号整数范围内。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def averageOfLevels(self, root: TreeNode) -> List[float]:
res = []
q = collections.deque([root])
while q:
n = len(q)
s = 0
for _ in range(n):
node = q.popleft()
s += node.val
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
res.append(s / n)
return res
/**
* 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) {
List<Double> res = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
while (!q.isEmpty()) {
double s = 0, n = q.size();
for (int i = 0; i < n; ++i) {
TreeNode node = q.poll();
s += node.val;
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
res.add(s / n);
}
return res;
}
}
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var averageOfLevels = function(root) {
let res = [];
let queue = [root];
while (queue.length > 0) {
n = queue.length;
let sum = 0;
for (let i = 0; i < n; i++) {
let node = queue.shift();
sum += node.val;
if (node.left) {
queue.push(node.left);
}
if (node.right) {
queue.push(node.right);
}
}
res.push(sum / n);
}
return res;
};