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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.
思路:跟level order一样,sum / size;
/**
* 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) {
List<Double> list = new ArrayList<Double>();
if(root == null) return list;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while(!queue.isEmpty()){
int size = queue.size();
Double levelSum = 0.0;
int step = size;
while(step>0){
TreeNode node = queue.poll();
step--;
levelSum += node.val;
if(node.left!=null){
queue.add(node.left);
}
if(node.right!=null){
queue.add(node.right);
}
}
list.add(levelSum / size);
}
return list;
}
}