1. The problem is to analyze the time complexity of the canonical encoding algorithm for a tree with $n$ nodes. 2. The given complexity is $O(n \log n)$, derived from sorting the children at each node. 3. The cost at node $i$ with $d_i$ children is $O(d_i \log d_i)$. 4. Summing over all nodes: $$T(n) = O\left(\sum_{i=1}^n d_i \log d_i\right)$$ 5. Since $\log d_i \leq \log n$, we have: $$T(n) \leq O\left(\sum_{i=1}^n d_i \log n\right) = O\left(\log n \sum_{i=1}^n d_i\right)$$ 6. In a tree, the sum of degrees is $\sum_{i=1}^n d_i = 2(n-1)$. 7. Therefore, $$T(n) = O(2(n-1) \log n) = O(n \log n)$$ ignoring constants. 8. For a more complex calculation, consider the exact distribution of $d_i$ values rather than bounding $\log d_i$ by $\log n$. 9. The exact sum is $$\sum_{i=1}^n d_i \log d_i$$ which depends on the degree distribution. 10. If degrees are uneven, e.g., one node with degree close to $n$ and others small, the complexity can approach $O(n \log n)$. 11. If degrees are balanced, the sum may be lower, but asymptotically still $O(n \log n)$. 12. Thus, the $O(n \log n)$ bound is tight and reflects the worst-case scenario. Final answer: The time complexity is $O(n \log n)$, and more complex calculations involve analyzing the degree distribution but do not improve the asymptotic bound.