given n identical values.
For Example:
For one node, the tree is a trivial one node tree.
For two nodes, two combinations are possible:
O O
\ and /
O O
For three nodes :
O O O O O
/ / / \ \ \
O O O O O O
/ \ \ /
O O O O
Answer:
/*
For the above diagrams, we can see the recursive property,
i.e,
For given N nodes:
* One node becomes the root
* Divide the remaining N-1 nodes between left and right subtrees, and this can be
done as follows:
Left Subtree Right Subtree
0 N-1
1 N-2
2 N-3
... ...
N-1 0
Summing up the above combinations, to get the value of N,
The solution can be recursively defined as following:
{ 1 if n == 1 }
F(n) = { sigma( F(i) * F(n-i-1 ) where i = 0 to n-1 }
note that its F(i) * F(n-i-1) , product, not a sum because its a combinatorics problem,
If event A can be in m ways, and another event B can be done in n ways, then the total
possible events are (m x n)
Note that optimization can be achived by caching the results in the recursive call.
*/
#include <stdio.h>
#include <stdlib.h>
#include <vector>
using namespace std;
// the cache used to "cache the results in a recursive call"
vectorcache;
size_t TreeCombinations(size_t n)
{
if ( (n == 1) || (n == 0) )
return cache[n];
// Check if the result was already computed
if ( cache[n] != 0 )
return cache[n];
else
{
int i = 0;
size_t combinations = 0;
//printf("{ %d \n", n);
for ( i =0 ; i <= n-1; ++i)
{
size_t tmp = TreeCombinations(i) * TreeCombinations(n-i-1);
//printf(" f(%d) * f(%d) = %d \n", i, n-i-1, tmp );
combinations += tmp;
}
//printf("}\n");
cache[n] = combinations;
}
return cache[n];
}
int main(int argc, char* argv[])
{
size_t n = atoi(argv[1]);
// initialize the cache
cache.reserve(n+1);
for ( int i = 0; i <= n; ++i )
{
cache[i] = 0;
}
cache[0] = 1;
cache[1] = 1;
// ----
return printf("Tree combinations for %d nodes = %d \n", n, TreeCombinations(n) );
}
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