diff --git a/problems/0222.完全二叉树的节点个数.md b/problems/0222.完全二叉树的节点个数.md
index ba7acc5a..746d45cc 100644
--- a/problems/0222.完全二叉树的节点个数.md
+++ b/problems/0222.完全二叉树的节点个数.md
@@ -646,5 +646,68 @@ func countNodes(_ root: TreeNode?) -> Int {
 }
 ```
 
+## Scala
+
+递归:
+```scala
+object Solution {
+  def countNodes(root: TreeNode): Int = {
+    if(root == null) return 0
+    1 + countNodes(root.left) + countNodes(root.right)
+  }
+}
+```
+
+层序遍历:
+```scala
+object Solution {
+  import scala.collection.mutable
+  def countNodes(root: TreeNode): Int = {
+    if (root == null) return 0
+    val queue = mutable.Queue[TreeNode]()
+    var node = 0
+    queue.enqueue(root)
+    while (!queue.isEmpty) {
+      val len = queue.size
+      for (i <- 0 until len) {
+        node += 1
+        val curNode = queue.dequeue()
+        if (curNode.left != null) queue.enqueue(curNode.left)
+        if (curNode.right != null) queue.enqueue(curNode.right)
+      }
+    }
+    node
+  }
+}
+```
+
+利用完全二叉树性质:
+```scala
+object Solution {
+  def countNodes(root: TreeNode): Int = {
+    if (root == null) return 0
+    var leftNode = root.left
+    var rightNode = root.right
+    // 向左向右往下探
+    var leftDepth = 0
+    while (leftNode != null) {
+      leftDepth += 1
+      leftNode = leftNode.left
+    }
+    var rightDepth = 0
+    while (rightNode != null) {
+      rightDepth += 1
+      rightNode = rightNode.right
+    }
+    // 如果相等就是一个满二叉树
+    if (leftDepth == rightDepth) {
+      return (2 << leftDepth) - 1
+    }
+    // 如果不相等就不是一个完全二叉树，继续向下递归
+    countNodes(root.left) + countNodes(root.right) + 1
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>