From 33ed3d980eecec6824fb5b9edad83d7b916c4142 Mon Sep 17 00:00:00 2001
From: ZongqinWang <1722249371@qq.com>
Date: Thu, 19 May 2022 10:35:55 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20102.=E4=BA=8C=E5=8F=89?=
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---
 problems/0102.二叉树的层序遍历.md | 25 +++++++++++++++++++++++
 1 file changed, 25 insertions(+)

diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index ab8f2e57..5afce73a 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -298,6 +298,31 @@ func levelOrder(_ root: TreeNode?) -> [[Int]] {
     return result
 }
 ```
+Scala:
+```scala
+// 102.二叉树的层序遍历
+object Solution {
+  import scala.collection.mutable
+  def levelOrder(root: TreeNode): List[List[Int]] = {
+    val res = mutable.ListBuffer[List[Int]]()
+    if (root == null) return res.toList
+    val queue = mutable.Queue[TreeNode]() // 声明一个队列
+    queue.enqueue(root) // 把根节点加入queue
+    while (!queue.isEmpty) {
+      val tmp = mutable.ListBuffer[Int]()
+      val len = queue.size // 求出len的长度
+      for (i <- 0 until len) { // 从0到当前队列长度的所有节点都加入到结果集
+        val curNode = queue.dequeue()
+        tmp.append(curNode.value)
+        if (curNode.left != null) queue.enqueue(curNode.left)
+        if (curNode.right != null) queue.enqueue(curNode.right)
+      }
+      res.append(tmp.toList)
+    }
+    res.toList
+  }
+}
+```
 
 **此时我们就掌握了二叉树的层序遍历了，那么如下九道力扣上的题目，只需要修改模板的两三行代码（不能再多了），便可打倒！**