(2^(N+1))-1
Suppose level is 2 then total number of nodes will be
1 root
2 left of root and right of root
2 left and right of left of root
2 left and right of right of root
so total nodes are 1+2+2+2=7
if the tree is binary tree [i.e two children max]then
2pow(N) is the answer
if it has 3 max children then
3pow(N)
...............................
if it has n max children then
npow(N)
to be more generic this kind of problem is best solved
recursivly.
To point out that 2 ^ N AND 3 ^ N are both wrong,
here's a few examples: (the exponet is the amount of levels)
2^0 = 1, correct
2^1 = 2, incorrect, should be 3
2^2 = 4, incorrect, should be 7
And a tree with three children
3^0 = 1, correct
3^1 = 3, incorrect, should be 4
3^2 = 9, incorrect, should be 13
Looking at that I'm sure you can see the pattern.
Let
C = "Number of Possible Children"
N = Levels
N
Σ C^N
j=0
or in C++ code
int NodeCount(int C, int N)
{
if (N < 0) return 0
return NodeCount(C, N-1) + C^N
}
along with your userid / name. ThanQ
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