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recurrence relation for binary tree height algorithm?

what would the recurrence relation be for the following pseudo code algorithm?

void Height(T)
{
     if (T is not empty)
     {
          lefttreeht = Height(left subtree of T);
          righttreeht = Height(right subtree of T);
         
          if (lefttreeht > righttreeht)
               return lefttreeht + 1;
          else
               return righttreeht + 1;
     }
     else
          return 0;
}

0
melissak
Asked:
melissak
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1 Solution
 
arnondCommented:
do you mean big O complexity ?
if so, it's O(log n), n being the number of nodes in the tree.

Arnon David.
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melissakAuthor Commented:
No, I mean;

     base case C(0)=0

recurrence relation for a tree of n nodes:

     C(n) = C(n-1) + something or nothing
     C(n+1) = C(n) + something or nothing

*parenthesis indicate subscript

this is what I think the format should somewhat be but
it could be something different.  I do know that the
explicit formula should be:

     C(n)=log(2)[n+1]-1

which was developed from some recurrence relation.
thanks
mk
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ozoCommented:
Are you assuming the tree is balanced?
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melissakAuthor Commented:
I don't think it matters.  The definition for the height of a
tree is: the number of nodes on the longest path from the root
to a leaf.  So, worst case binary tree would be one that has nodes attatched to the left or right side only (a diagonal).
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ozoCommented:
so is C(n) supposed to be the worst case height of a balanced tree of n nodes?
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heyhey_Commented:
maybe i am missing something, but if
the explicit formula should be:
     C(n)=log(2)[n+1]-1
than the recurrence relation should be
     C(n+1)=C(n) + log(2)[n+1]
     
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gugaCommented:
I have one comment
if C(n)=log(2)[n+1]-1 then let's n=4

C(n)=log(2)[5]-1;

but height of tree must be integer!
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gugaCommented:
may be (i don't sure)
c(n+1) = C(n) + 1
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melissakAuthor Commented:
I had math student help me.  With each call to the algorithm the list of N elements is halved.  Since the algorithm recursively calls itself twice the formula is:
C(N) = 2*C(N/2)
To Prove relation:
Let N = 2^n

C(2^n) = 2*C(2^n-1)
C(2^n-1) = 2*C(2^n-2)
C(2^n) = 2^k * C(2^n-k)
Let k=n
C(2^n) = 2^n * C(2^0) = 2^n

since 2^n = N

C(N) = N

BigOH = O(N)

sorry took so long to respond.  e-mail and internet down.
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