# 102. Binary Tree Level Order Traversal (BFS)
BFS Traversal
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問題點
**在每一層遍歷開始前，先記錄隊列中的結點數量 n**
```java
int levelNums = queue.size();
```
所以需要利用當前 queue 的 size 當分層的數量
```java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List> levelOrder(TreeNode root) {
Queue queue = new LinkedList<>();
List> result = new ArrayList<>();
if (root == null) return result;
queue.offer(root);
while (!queue.isEmpty()) {
int levelNums = queue.size();
List subList = new ArrayList<>();
for (int i = 0; i < levelNums; i++) {
TreeNode node = queue.poll();
subList.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
result.add(subList);
}
return result;
}
}
```
但如果不是樹的題目
因為不是樹, 所以不用 queue size 來分層遍歷