just count it? yes AC...
But we need to use the prpoerties of complete tree
except the last level of the tree other level is full which means we only need to count the last'nodes
we can use this properties
in my way 2^n-1 + last level's nodes is a great solution
however!!! the last level's nodes is hard to summarizing.
Therefore the std use another method to implement this properties.
It summary whether current nodes have left most and right most. If it has it must equal to the 2^h
else just take the left and right as the new tree!