add scala #129

scala-leetcode/README.md

@@ -65,6 +65,7 @@ Leetcode Scala 实现
 * [填充每个节点的下一个右侧节点指针](./src/main/scala/io/github/dreamylost/Leetcode_116.scala)
 * [填充每个节点的下一个右侧节点指针 II](./src/main/scala/io/github/dreamylost/Leetcode_117.scala)
 * [分裂二叉树的最大乘积](./src/main/scala/io/github/dreamylost/Leetcode_1339.scala)
+* [求根到叶子节点数字之和](./src/main/scala/io/github/dreamylost/Leetcode_129.scala)
 
 ## 链表
 

scala-leetcode/src/main/scala/io/github/dreamylost/Leetcode_129.scala

+/* Licensed under Apache-2.0 @梦境迷离 */
+package io.github.dreamylost
+
+/**
+  * 129. 求根到叶子节点数字之和
+  *
+  * 给定一个二叉树，它的每个结点都存放一个 0-9 的数字，每条从根到叶子节点的路径都代表一个数字。
+  *
+  * 例如，从根到叶子节点路径 1->2->3 代表数字 123。
+  *
+  * 计算从根到叶子节点生成的所有数字之和。
+  *
+  * 说明: 叶子节点是指没有子节点的节点。
+  *
+  * @author 梦境迷离 dreamylost
+  * @since 2020-07-16
+  * @version v1.0
+  */
+object Leetcode_129 extends App {
+
+  val ret = sumNumbers(TreeNodeData.treeData5_1())
+  println(ret)
+
+  /**
+    * 620 ms,100.00%
+    * 51.2 MB,100.00%
+    * 最优解是：利用进制
+    *
+    * @see Leetcode_1022
+    * @param root
+    * @return
+    */
+  def sumNumbers(root: TreeNode): Int = {
+    var ret = Seq.empty[String]
+
+    def helper(r: TreeNode, path: String): Unit = {
+      if (r == null) return
+      if (r.left == null && r.right == null) {
+        ret = ret ++ Seq(path + r.value.toString)
+      }
+      helper(r.left, path + r.value.toString)
+      helper(r.right, path + r.value.toString)
+    }
+
+    helper(root, new String)
+    ret.map(p => Integer.parseInt(p)).sum
+  }
+}