Today we continue more coding puzzles reviews. One of the problems is about serializing and deserializing a binary tree.
There are many ways to do it. Usually two different orders of traversal of a binary tree can give all the information to deserialize the binary tree. They are also efficient because each node is visited once during a traversal. Another way to do this would be to print the tree level wise with null nodes for missing child nodes of the tree. The level wise ensures one serialization result.
void serialize(Node root, Node delimiter, Node Empty)
{
var q = new Queue<Node>();
q.Enqueue(root);
q.Enqueue(delimiter);

var node = q.Dequeue();
while (node)
{
if (root.left)
      q.Enqueue(root.left) ;
else
      q.Enqueue(Empty); // new node

if (root.right)
      q.Enqueue(root.right);
else
      q.Enqueue(Empty); // new node

node = q.Dequeue();

Console.Write(node.toString());

if (node == delimiter) {q.Enqueue(delimiter);}

}
}

The deserialization now will find all the left or child nodes for every parent nodes even if they are null.

void deserialize(List<Node> nodes, Node delimiter)
{
int level = nodes.Index(nodes.First(x => x == delimiter));
level++;
for (int I = 0; I < nodes.Count() && level < nodes.Count(); I++)
{
   if (nodes[I] != delimiter)
   { nodes[I].left = nodes[level]; level++;
      nodes[I].right = nodes[level]; level++;
   }
   else
        assert nodes[level] ==delimiter
        level++;
}
}

Let's take an example:
   1
2   3
    4  5
1D23DEE45