Java solution 257 & 258

Author: gavinfish

java/_257BinaryTreePaths.java

@@ -0,0 +1,41 @@
+import java.util.List;
+
+import javax.management.relation.RelationTypeSupport;
+import javax.swing.tree.TreeNode;
+
+/**
+ * Given a binary tree, return all root-to-leaf paths.
+ * 
+ * For example, given the following binary tree:
+ * 
+ *    1
+ *  /   \
+ * 2     3
+ *  \
+ *   5
+ * All root-to-leaf paths are:
+ * 
+ * ["1->2->5", "1->3"]
+ */
+
+public class _257BinaryTreePaths {
+    public List<String> binaryTreePaths(TreeNode root) {
+        List<String> result = new ArrayList<>();
+        if (root != null) {
+            search(root, "", result);
+        }
+        return result;
+    }
+
+    public void search(TreeNode node, String path, List<String> result) {
+        if (node.left == null && node.right == null) {
+            result.add(path + node.val);
+        }
+        if (node.left != null) {
+            search(node.left, path + node.val + "->", result);
+        }
+        if (node.right != null) {
+            search(node.right, path + node.val + "->", result);
+        }
+    }
+}

java/_258AddDigits.java

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+/**
+ * Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
+ * 
+ * For example:
+ * 
+ * Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
+ * 
+ * Follow up:
+ * Could you do it without any loop/recursion in O(1) runtime?
+ */
+public class _258AddDigits {
+    public int addDigits(int num) {
+        int result = num % 9;
+        return (result != 0 || num == 0) ? result : 9;
+    }
+}