Ancestor Check

Node $u$ is an ancestor of $v$ if DFS entered $u$ before $v$ and exited $u$ after $v$. Condition: tin[u] <= tin[v] and tout[v] <= tout[u]. Check both inequalities in $O(1)$ time. This is faster than LCA for simple ancestor queries. No extra data structures needed, $tin$ and $tout$ arrays.

This property makes it easy to check relationships between nodes without walking the tree. This check runs in $O(1)$ time. No tree traversal needed. Compare two pairs of numbers and you get the answer.

Space complexity is $O(n)$ for the data structures used.