题目及思路
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.
取每层的平均值，则是层序遍历（BFS）
代码
public class AverageOfLevelsInBinaryTree {
public static List<Double> averageOfLevels(TreeNode root) {
List<Double> averageList = new ArrayList<>();
List<Integer> tmpList = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
if (root == null) {
return null;
}
queue.add(root);
while (!queue.isEmpty()) {
long sum = 0;//注意得是long型，否则会溢出
double avarage = 0;
int n = queue.size();
for (int i = 0; i < n; i++) {
TreeNode currentNode = queue.peek();
sum += currentNode.val;
queue.poll();
if (currentNode.left != null) {
queue.offer(currentNode.left);
}
if (currentNode.right != null) {
queue.offer(currentNode.right);
}
}
avarage = (double)sum / n;
averageList.add(avarage);
}
return averageList;
}
