$$$timein[x]$$$ is when you enter $$$x$$$ and $$$timeout[x]$$$ is when you leave $$$x$$$, these values can be calculated as below:

t = 0;
void dfs(u){
   timein[u] = ++t;
   for(int v: adj[u])
      if(!vis[v])
          dfs(v)
   timeout[u] = ++t;
}

If $$$a$$$ is an ancestor of $$$b$$$ only if this inequality is satisfied: $$$timein[a] ≤ timein[b] ≤ timeout[a]$$$. To check if some node $$$x$$$ presents in the simple path from root to $$$a$$$, it is enough to check if the root is an ancestor of $$$x$$$ and $$$x$$$ is an ancestor of $$$a$$$. We have to pre process in $$$O(n)$$$ then we can answer query in $$$O(1)$$$.

Could a O(logn) solution be that if LCA of target node and the given node A is target node itself, it is present. Else, it is not?