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.
分析¶
这道题目非常显然的可以用广度优先搜索。搜索每一层,然后计算该层的平均值。
public List<Double> averageOfLevels(TreeNode root) { List<Double> average = new ArrayList<>(); if (root == null) return average; Queue<TreeNode> queue = new LinkedList<>(); queue.off(root); while (!list.isEmpty()) { long sum = 0; int size = list.size(); for (int i = 0; i < size; i++) { TreeNode node = queue.poll(); sum += node.val; if (node.left != null) queue.offer(node.left); if (node.right != null) queue.offer(node.right); } average.add(((double)sum)/size); } return average; }





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