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.
Solution:
/**
* 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<>();
List<Integer> list = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while(!queue.isEmpty())
{
int size=queue.size();
for(int i=0;i<size;i++)
{
TreeNode current=queue.poll();
list.add(current.val);
if(current.left!=null) queue.add(current.left);
if(current.right!=null) queue.add(current.right);
}
double sum=0.0,avg=0.0;
for(int i:list)
{
sum+=i;
}
avg=sum/list.size();
result.add(avg);
list.clear();
}
return result;
}
}
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