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.
M1: BFS
注意!sum要用long,用int会溢出
time: O(n), space: O(N) -- 最多一层节点的个数
/** * 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> res = new ArrayList<>(); if(root == null) { return res; } Queue<TreeNode> q = new LinkedList<>(); q.offer(root); while(!q.isEmpty()) { long sum = 0; int size = q.size(); for(int i = 0; i < size; i++) { TreeNode t = q.poll(); sum += t.val; if(t.left != null) { q.offer(t.left); } if(t.right != null) { q.offer(t.right); } } res.add((double)sum / size); } return res; } }
M2: DFS
time: O(n), space: O(height)