Size of a serialized complete binary tree - serialization

Question

I'm tryin to work out the size of a serialized binary tree having N nodes (also mentioned in Leetcode). This is how I calculate the size:
If we assume the storage required to store values be V bits for each node, then the storage needed to store N nodes will be N.V. We also need to store NULL for the leaves; since there are exactly Ceiling(N/2) leaves in a complete tree, and assuming only one bit is enough to represent NULL, then an additional of 2 x Ceiling(N/2) bits will be required. 2 x Ceiling(N/2) translates to N+1 as in a complete tree N is always an odd number.
So, N.V + (N+1) bit is required in total.
However, I can see that in Leetcode and some other places (e.g. this), it's calculated as N.V + 2N.
What am I missing?

Answer 1

What am I missing?
The two references you provided (LeetCode and blog article) deal with arbitrary binary trees, not necessarily complete. So let me first deal with arbitrary binary trees:
Although a NULL reference could be represented with one bit (e.g. with value 0), you also need to store the fact that a reference is not a NULL (value 1). You cannot just omit the bit, as then the next bit (belonging to a node value) could be misinterpreted as indicating a NULL reference. So you should not only count that bit for each NULL reference, but count it in for all branches.
The serialised format would for each node represent:
The node's value (𝑉 bits)
The fact whether or not its left child is a NULL (1 bit)
The fact whether or not its right child is a NULL (1 bit)
Example:
Let 𝑉 be 4
Tree to serialise:
10
/ \
7 13
\
14
Serialisation process (level order):
node value
has left
has right
serialised
without spacing
10
yes
yes
1010 1 1
101011
7
no
no
0111 0 0
011100
13
no
yes
1101 0 1
110101
14
no
no
1110 0 0
111000
Complete:
101011011100110101111000
If we were only to store the 0 when there is a NULL, then we would get this:
101001110011010111000
^
But now the bit at the indicated position is ambiguous, because that bit could be interpreted as representing a NULL reference, but actually it is the first of 𝑉 bits 0111 representing the value 7.
It is however possible to reduce the serialised string with 2 bits: the very last 2 bits will always be 0 in a tree traversal that is guaranteed to end with a leaf. This is for example the case for level-order and pre-order traversals. So then you could just omit those 2 bits.
The case for complete binary trees
First of all about the definition of a complete binary tree. You write:
in a complete tree N is always an odd number.
I suppose then your definition of a complete tree is what at Wikipedia is called a perfect tree. We can however also look at (nearly) complete binary trees (and then 𝑁 is not necessarily odd).
For complete binary trees the case is simpler, as a level order traversal of a complete binary tree will never include NULLs, i.e. there are no "gaps" in such a traversal.
So you can just serialise the node's values in that order, giving each 𝑉 bits. This is actually the array representation that is used for binary heaps:
The parent / child relationship is defined implicitly by the elements' indices in the array.
If serialisation happens in a string data type that implicitly has a length attribute, then that's it. If there is no such meta data, then you need to prefix the value of 𝑁 in the serialisation, reserving a predefined number of bits for it. Alternatively, if there is a special value of 𝑉 bits that will never occur as actual node value, you could append it as a terminator (much like \0 in C-strings).