Time analysis of binary search tree operations

I read about binary search trees that if it is a complete tree (all nodes except leaf nodes have two children) having n nodes, then no path can have more than 1+log n nodes.

Here is the calculation I did... can you show me where did I go wrong....

the first level of bst has only one node(i.e. the root)-->2^0
the second level have 2 nodes(the children of root)---->2^1
the third level has 2^3=8 nodes
 .
 .
the (x+1)th level has 2^x nodes

so开发者_StackOverflow中文版 the total number of nodes =n = 2^0 +2^1 +2^2 +...+2^x = 2^(x+1)-1
so, x=log(n+1)-1

now as it is a 'complete' tree...the longest path(which has most no of nodes)=x
and so the nodes experienced in this path is x+1= log(n+1)

Then how did the number 1+log n come up...?

Shorter answer: the number x of levels in a complete (or perfect) binary tree is log2(n+1), where n is the number of nodes (alternatively, n = 2^(x-1)). A tree with x levels has height x-1. The longest path from the root to any node contains x = log2(n+1) nodes (and x-1 edges).

Now because n+1 is a power of 2, we have that log2(n+1) = 1 + floor(log2(n)). In other words, 1 + log2(n) is a correct upper-bound, but it is never an integer.

It is unclear to me whether the x in your computation refers to the height or the number of levels.