Added question 297.

Author: luceCoding

Diff for: leetcode/hard/297_serialize_and_deserialize_binary_tree.md

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+# 297. Serialize and Deserialize Binary Tree
+
+## DFS Solution
+- Run-time: O(N)
+- Space: O(N)
+- N = Number of nodes in tree
+
+We cannot reuse the solutions from question #105. or #106 using two traversals because duplicate values are allowed.
+We would need a different method.
+
+A DFS method is probably the easiest method to come up with.
+I chose to use preorder, you can implement this using any of the traversals.
+
+Let's start with serialize(), we can simply traverse via. preorder and construct a list.
+This part is straightforward, however, there is a caveat and that is related to space and how we should represent None nodes.
+We could use a string like 'None' but we can save more space by using an empty string instead, this would require us to use a delimiter.
+
+For deserialize(), using the delimiters, we can get the list of nodes in preorder.
+It would simply be a preorder traversal to reconstruct the tree.
+
+There are some further optimizations that we could've done, like condensing consecutive empty strings together.
+It would be more over engineering at this point but good to mention.
+
+```
+class Codec:
+
+    curr_idx = 0
+    def serialize(self, root):
+        """Encodes a tree to a single string.
+
+        :type root: TreeNode
+        :rtype: str
+        """
+        def preorder_encode(root):
+            if root is None:
+                result.append('')
+                return
+            result.append(str(root.val))
+            preorder_encode(root.left)
+            preorder_encode(root.right)
+
+        result = list()
+        preorder_encode(root)
+        return ','.join(result)
+
+    def deserialize(self, data):
+        """Decodes your encoded data to tree.
+
+        :type data: str
+        :rtype: TreeNode
+        """
+        def preorder_decode():
+            if self.curr_idx > len(tokens) or tokens[self.curr_idx] == '':
+                return None
+            root = TreeNode(int(tokens[self.curr_idx]))
+            self.curr_idx += 1
+            root.left = preorder_decode()
+            self.curr_idx += 1
+            root.right = preorder_decode()
+            return root
+
+        tokens = data.split(',')
+        return preorder_decode()
+```