# how do i check to see if a binary tree is complete

i have a binary tree.  I can count the nodes,add,delete, and check to see if it is full, but what i cannot figure out is how to check to see if it is complete
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Commented:
See the definition at http://www.nist.gov/dads/HTML/completeBinaryTree.html - you have to check whether this requirement is met:

"A complete binary tree has 2k nodes at every depth k < n and between 2n and 2n+1-1 nodes altogether"
Author Commented:
That i understand. and i also know that i should do it recursively but i cannot figure out code or even psydo code to check this
Commented:
What is the definition of a complete binary tree? One that is full down to heigh - 1, right? So you can check for that with a modified full check (one that takes a max depth to check). The other requirement is harder: all nodes in the bottom row are as far to the left as possible.

Note that this problem is really only hard if you are storing your binary tree using pointers in the traditional manner. If you store your binary tree in an array it is easy to check the index of the last child and the number of children. They will match in a complete tree.

So, back to the hard one:  The left subtree must be full (if there are enough leaves in the bottom tier) or it must be complete. The left subtree must be full (if there are too few nodes to spill over in the last row) or it must be complete. Thus we have a recursive relationship between a complete tree and its left and right subtrees.

Yeah!

Hope this helps (ask if you don't understand what I said).

-bcl
Author Commented:
I think i understand what your saying..but do u start checking from the beginning or from the bottom and and how would you do that recursively?
Commented:

(1) The recursive relationship I described above holds from the root down. That is you call either full or complete on the left and right subtrees and return the and of the two values.

(2) I thought of an alternative algorithm:
do a breadth first traversal of the whole tree.
store an integer in each node in the tree indicating the order in which it was visited
in the depth first traversal (root node is 0).

do an in-order (order doesn't matter here) traversal and check at each node that:
both subtrees are empty or the right subtree is empty
if a subtree is not empty:
if its a left subtree the node number should be 2 * n + 1 where n is the number of the current node
if it is a right subtree the node number should be 2 * n + 2.

Maps the array storage mechanism onto the pointer based binary tree.

Hope this helps, -bcl
Author Commented:
I like the 1 recursive idea.  But i dont fully understand it. i understand that you check each not to see weather or not it has 2 children. would you return false if it doesnt have 2 children and its not the next to last row?(i do have a function to count the height so that should help)
Commented:
The following is not the most effecient algorithm in the universe (but I can think of several ways to speed it up by caching results of some of the funcitons):

complete(nodePtr curr) {
if (!curr) return true;
n = sizeOfTree(curr);
h = heightOfTree(curr);
if (h > floor(logbase2(n)) + 1) return false; // too tall to work
// If it IS a complete tree we know the bottom row must be populated from the left
// the question is where do those nodes end? Do they go more than halfway across?
// How would we know if they went more than halfway across? Well we can remove the
// nodes in the "full" part of the tree from consideration and then see:
f = nodesInAFullTree(h-1);
k = n - f; // nodes in the last row

if (k > f / 2)  // why?
return full(curr->left) && complete(curr->right);
else
return complete(curr->left) && full(curr->right);

(Note I haven't compiled and tested this code and there could be off by 1 errors in the comparisons).

-bcl

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Commented:
Why a C? I provided two different approaches to the problem.

-bcl
Commented:
And both algorithms actually work. (Just wrote a quick test.)

Okay, there are a couple of missing checks in the recursive version:

return false if
- size of left subtree is less than size of right subtree
- height of left subtree is less than height of right subtree
- height of left subtree is more than one greater than height of right subtree

With those tests in the two find all of the complete trees for tree sizes up to 12 nodes (I ran them across all permutations of insertion order into the bst; not the most efficient testing regime but it is exhaustive).

So, again, what is the deal with posting the question again and giving my a lousy grade? I believe I answered the question.

-bcl
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