Why the root of this tree has to be "1"? Arrange $2^{n-1}-1$ zeroes and $2^{n-1}$ ones in a balanced full binary tree of depth $n$. If we want the number of edges that connect the same (and respectively different) digits are the same, then one claims that the root of this tree has to be a one. Why is that?
For example, if $n=2$, then we need to arrange 1 zero and 2 ones. One arrangement that makes the number of edges that connect the same digits (which in this case is only one: the edge with a "+" who connects 2 ones,) and the number of edges that connect two different digits (which in this case is also one: the edge with a "-") are the same. Note that the root of this tree is unexceptionally 1.
      1
    /   \
 + /     \ -
  /       \
 1         0

And here is a case where $n=3$.
          1
        /   \
     + /     \ -
      /       \
     1         0
  + / \ -   + / \ -
   /   \     /   \
  1     0   0     1

Again, we see that the root is a one. If we change the root to a zero, however, then we can never find an arrangement that makes the number of "+" edges equals to the number of "-" edges. Why is that?
 A: Consider a balanced depth-$n$ binary tree of $0$'s and $1$'s and label each edge $+$ or $-$ as you did.
If  all edges were $+$, all $2^n-1$ vertices would be the same as the root.  You can switch
an edge from $+$ to $-$ or vice versa if you flip all the vertices above that edge (i.e. those such that the edge is on the path from the vertex to the root).  This always involves
flipping an odd number of vertices, so the parity of the numbers of $1$'s and of $0$'s changes.  If you do an even number of edge-switches, you have the same parity you started with.  Now your tree has $2^n-2$ edges, and you need to do $2^{n-1}-1$ switches to get an equal number of $+$ and $-$ edges.  If $n \ge 2$, this is odd, so the parity is  different than at the start.  At the start, there were an odd number the same as the root, so after the switching you must have an even number.  Since you want the number of $0$'s to be odd,
the root must be $1$.
A: Let's say you have $k$ edge incidences at $1$s and $l$ edge incidences at $0$s. Then if $x$ edges connect a $0$ and a $1$, that leaves $k-x$ incidences at $1$s, so there are $(k-x)/2$ edges connecting a $1$ and a $1$, and likewise $(l-x)/2$ edges connecting a $0$ and a $0$. Then the condition that there are the same number of $+$ and $-$ edges is
$$
\frac{k-x}2+\frac{l-x}2=x\;,
$$
and thus $x=(k+l)/4$. Thus there are
$$
\frac{k-x}2=\frac{k-(k+l)/4}2=\frac38k-\frac18l
$$
edges connecting a $1$ and a $1$. Swapping a $0$ and a $1$ between a branch and a leaf changes this quantity by $\pm1$, so the only way to change whether it's an integer is to change the root, and it turns out that it's an integer if the root is a $1$ and not an integer if it isn't.
