题目
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.
讲解
这题跟429. N-ary Tree Level Order Traversal很像。都是广度优先遍历树(也就是按层级遍历树)。
java代码
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
List<Double> result = new ArrayList<>();
public List<Double> averageOfLevels(TreeNode root) {
if(root==null){
return result;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
int children_num=1;
while(!queue.isEmpty()){
double sum = 0;
int count=0;
for(int i=0;i<children_num;i++){
TreeNode now = queue.poll();
sum += now.val;
if(now.left!=null){
count++;
queue.offer(now.left);
}
if(now.right!=null){
count++;
queue.offer(now.right);
}
}
sum /= children_num;
result.add(sum);
children_num = count;
}
return result;
}
}
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