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.
代码
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Double> averageOfLevels(TreeNode root) {
ArrayList<Double> list = new ArrayList<Double>();
LinkedList<TreeNode> tree = new LinkedList<TreeNode>();
tree.add(root);
double sum = 0;
while(!tree.isEmpty()){
int n = tree.size();
sum = 0;
for(int i=0;i<n;i++){
TreeNode node = tree.removeFirst();
sum+=node.val;
if(node.left!=null){
tree.addLast(node.left);
}
if(node.right!=null){
tree.addLast(node.right);
}
}
list.add(sum/n);
}
return list;
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Double> averageOfLevels(TreeNode root) {
ArrayList<Double> list = new ArrayList<Double>();
LinkedList<TreeNode> tree = new LinkedList<TreeNode>();
tree.add(root);
double sum = 0;
while(!tree.isEmpty()){
int n = tree.size();
sum = 0;
for(int i=0;i<n;i++){
TreeNode node = tree.removeFirst();
sum+=node.val;
if(node.left!=null){
tree.addLast(node.left);
}
if(node.right!=null){
tree.addLast(node.right);
}
}
list.add(sum/n);
}
return list;
}
}
