题目链接
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。
示例 1:
输入:root = [3,9,20,null,null,15,7]输出:[3.00000,14.50000,11.00000]解释:第 0 层的平均值为 3,第 1 层的平均值为 14.5,第 2 层的平均值为 11 。因此返回 [3, 14.5, 11] 。
示例 2:
输入:root = [3,9,20,15,7]
输出:[3.00000,14.50000,11.00000]
提示:
- 树中节点数量在 [1, 104] 范围内
- -231 <= Node.val <= 231 - 1
/**
* 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> result = new ArrayList<>();
if (root == null) {
return result;
}
Deque<TreeNode> deque = new LinkedList<>();
deque.offer(root);
while (!deque.isEmpty()) {
int size = deque.size();
double sum = 0.0;
for (int i = 0; i < size; i++) {
TreeNode temp = deque.peek();
deque.pop();
sum += temp.val;
if (temp.left!= null) {
deque.offer(temp.left);
}
if(temp.right != null) {
deque.offer(temp.right);
}
}
result.add(sum / size);
}
return result;
}
}
