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Source. Sections 4.0–4.2, printed p. 454 (published PDF).

Statement. An alternating tree with inner set $I$ and outer set $O$ has $|O|=|I|+1$. Its maximum matching size is $|I|$. For each $v\in O$ there is a unique maximum matching leaving $v$ exposed, and every maximum matching arises in this way.

Proof. Counting edges by their inner endpoints gives $|E(T)|=2|I|$. Since a finite tree has one fewer edge than vertices, $2|I|=|I|+|O|-1$, proving the count. Every matching edge uses a different inner vertex, so its size is at most $|I|$.

We prove existence and uniqueness for a prescribed $v$ by induction on $|I|$. For $|I|=0$, the tree is the singleton $v$ and the empty matching is unique. Otherwise choose an inner vertex $u$, with outer neighbors $a,b$. Deleting $u$ separates the tree into two alternating trees, one containing $v$. Name that component $T_a$, so its attachment neighbor is $a$, and call the other $T_b$. By induction, $T_a$ has a unique matching omitting $v$, and $T_b$ one omitting $b$. Together with $ub$, they match every vertex except $v$.

Conversely, any matching omitting only $v$ must match $u$. It cannot use $ua$: the odd component $T_b$ would then have no crossing matching edge and would need a perfect matching. It must therefore use $ub$. The restrictions to $T_a$ and $T_b$ omit respectively $v$ and $b$, so induction forces them uniquely. This supplies the parity step implicit in the source's induction.

The constructed matching has $|I|$ edges. Every maximum matching therefore covers all inner vertices and exactly $|I|$ outer vertices, leaving just one outer vertex exposed. $\square$