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What is the maximum total number of nodes in a tree that has
N levels? Note that the root is level (zero)
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Answer / salmiya thilsath.a
2^(N+1)-1..
if N=0; it is 2-1=1,1 is the max no of node in the tree
if N=1; it is 4-1=3, 3 is the max no of nodes in the tree
if N=2; it is 8-1=7, 7 is the max
and it goes like that...........
| Is This Answer Correct ? | 124 Yes | 8 No |
Answer / asharani k p
(2 pow n)-1,if root level is one and (2 pow n+1)-1, if root
level is zero.
| Is This Answer Correct ? | 26 Yes | 11 No |
Answer / gaurav gupta
(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
by formula (2^(2+1))-1
8-1=7
| Is This Answer Correct ? | 24 Yes | 10 No |
Answer / bipin from utkal university mc
if tree is binary tree then maximum no.of node is 2^(N+1)-1
| Is This Answer Correct ? | 5 Yes | 0 No |
Answer / hex
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
}
| Is This Answer Correct ? | 8 Yes | 4 No |
Answer / vaishali naidu
If root is at level 0 then :
Case 0:
When level is 1 max nodes is 1
Case 1:
When level is 1 then max would be 3.
Case 2:
When level is 2 then max nodes would be 7
So formula would be 2^(n+1) -1
2^(0+1)-1=1
2^(1+1)-1=3
2^(2+1)-1=7
| Is This Answer Correct ? | 6 Yes | 3 No |
Answer / s. sadagopan
Infinite
In a tree a node can have any number of nodes independent
of the levels.
| Is This Answer Correct ? | 8 Yes | 6 No |
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