Kraft inequality and effective prefix-code allocation

At depth $N$, a word $p$ excludes exactly $2^{N-\mathrm{\mid }p\mathrm{\mid }}$ leaves; prefix-freeness makes these excluded sets disjoint. Divide their total by $2^N$ and let $N$ grow to obtain the inequality.

For requests, maintain a finite prefix-free set $F$ of free tree nodes, initially $F=\{\epsilon \}$ for the empty word $\epsilon$. Require that its nodes have pairwise distinct lengths and that their cylinders are exactly the unallocated part of the binary tree. After finitely many requests, the total free capacity is ${\sum }_{\sigma \in F}{2}^{-\mathrm{\mid }\sigma \mathrm{\mid }}=1-{\sum }_{i\text{ already allocated}}{2}^{-n_i}$. For the next request of length $n$, this capacity is at least $2^{-n}$ by the bound on the entire request list. If every node of $F$ had length greater than $n$, distinctness of the lengths would make its capacity at most ${\sum }_{j>n}{2}^{-j}=2^{-n}$, with strict inequality because $F$ is finite. Therefore $F$ contains a node of length at most $n$. Choose the one $\sigma$ of greatest such length $m$; it is unique because the free-node lengths are distinct.

Remove $\sigma$ from $F$. Allocate the descendant $\sigma 0^{n-m}$ as the requested codeword, and insert the siblings $\sigma 0^{j}1$ for $0\le j<n-m$ into $F$. Their depths are $m+1,\dots ,n$; no existing free node has one of these depths, by maximality of $m$ among depths at most $n$. The new free nodes are prefix-free and partition exactly the part of $[\sigma ]$ outside the allocated cylinder, so both invariants and the capacity identity persist. Each step is a finite computable operation on the current finite set $F$. Processing the computable enumeration in order therefore gives distinct codewords of the requested lengths and a prefix-free effective allocation, including the empty-word case $n=0$.