Merge pull request #1391 from wzqwtt/tree08

添加（0222.完全二叉树的节点个数、0257.二叉树的所有路径）Scala版本

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
+  }
+}
+```
+
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0257.二叉树的所有路径.md

@@ -702,5 +702,35 @@ func binaryTreePaths(_ root: TreeNode?) -> [String] {
 }
 ```
 
+Scala:
+
+递归:
+```scala
+object Solution {
+  import scala.collection.mutable.ListBuffer
+  def binaryTreePaths(root: TreeNode): List[String] = {
+    val res = ListBuffer[String]()
+    def traversal(curNode: TreeNode, path: ListBuffer[Int]): Unit = {
+      path.append(curNode.value)
+      if (curNode.left == null && curNode.right == null) {
+        res.append(path.mkString("->")) // mkString函数: 将数组的所有值按照指定字符串拼接
+        return // 处理完可以直接return
+      }
+
+      if (curNode.left != null) {
+        traversal(curNode.left, path)
+        path.remove(path.size - 1)
+      }
+      if (curNode.right != null) {
+        traversal(curNode.right, path)
+        path.remove(path.size - 1)
+      }
+    }
+    traversal(root, ListBuffer[Int]())
+    res.toList
+  }
+}
+```
+
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>