Which of these two methods provides the correct answer for this probability riddle? First, I know this riddle has been asked (many times) before. The question I want answering is why is a tree diagram not a correct method for determining the probability in this case.

There are two children, equally likely to be Boy or Girl.   If we know
  one (or more) is a Boy, what is the probability that there is a Girl in the
  pair?

The sample space looks like this:
BB
BG
GB
GG - not possible
Therefore the probability is 2/3, but if I draw a tree diagram:

The probability of a girl in the pair seems to be $$\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$$
Why does a tree diagram fail?

Your probiabilities are not correct. Note that if we call the two children $X_1$ and $X_2$, then 
\begin{align*}
   \def\P{\mathbf P}\P[X_1 = B \mid X_1 = B \vee X_2 = B]
    &= \frac{\P[X_1 = B]}{\P[X_1 = B \vee X_2 = B]}\\
    &= \frac{\frac 12}{\frac 34}\\
    &= \frac 23
\end{align*}
So, the probiabilities at the first node should be $\frac 23$ and $\frac 13$, respectively, giving 
$$ \frac 23 \cdot \frac 12 + \frac 13 = \frac 13 + \frac 13 = \frac 23 $$
in total.
