Number of leaves in a tree (all types of trees)

Count the number of edges in two ways.

  • By the Handshake Lemma: $|E|=\frac{1}{2}\sum\limits_{i=1}^\infty i|V_i|$.
  • By the characterization of trees: $|E|= \sum\limits_{i=1}^\infty |V_i|-1$.

    Equate these two quantities, and remember that $|V_1|$ is the number of leaves.

    $\frac{1}{2}\sum\limits_{i=1}^\infty i|V_i|= \sum\limits_{i=1}^\infty |V_i|-1,$

    so

    $\sum\limits_{i=1}^\infty (i-2)|V_i|+2=0$

    so

    $-|V_1|+0+\sum\limits_{i=3}^\infty (i-2)|V_i|+2=0$

    so

    $|V_1|=\sum\limits_{i=3}^\infty (i-2)|V_i|+2$

    which is what you are after.


  • Prove it by induction on $n$:

    • It's trivial for the case $n=2$. (Why?)
    • The inductive step is to add a vertex and, because it's a tree, you've changed at most three of the $V_i$. (Which ones? How have they changed? Understand this before proceeding.) In the sum, then, you're adding $1$ to the left-hand side and $-(k-2)+(k+1-2)$ for some $k$ to the right, which maintains the equality.

    There is a close relationship between your problem and the degree sum formula for graphs. For any finite graph, tree or otherwise, the equation

    $$2|E| = \sum_{v\in V} \deg(v)$$

    always holds. If you haven't proved this before, it's worth stopping to think about why the equation is correct.

    It is possible to transform the equation

    $$|V_1| = 2 + \sum (j-2)|V_j|$$

    into an explicit statement of the degree sum theorem for trees. Hint: $j|V_j|$ is just another way to write $\sum_{v\in V_j} \deg(v)$.