Let $I$ represent the set of all interior vertices in a binary tree. Define the probability of vertex $v \in I$ is $P(v)$, which is the sum of the probabilities of all leaf vertices that are descendants of $v$. Show that (or given a counter-example)
$$L(C) = \sum_{v\in \mathcal{I}}^{} P(v) .$$
Suppose that $u$ is a leaf at depth $d(u)$; then there are $d(u)$ interior vertices in the path from $u$ to the root, so $u$ contributes $d(u)P(u)$ to $\sum_{v\in I}P(v)$. Thus, if $L$ is the set of leaves,
$$\sum_{v\in I}P(v)=\sum_{v\in L}d(v)P(v)\,.$$
On the other hand, $d(u)$ is the length of the code word corresponding to a leaf $u$, so $\sum_{v\in L}d(v)P(v)$ is the mean length of a code word of $C$.