# Finding an MST with one adding and removing vertex operation

I am facing the following problem: Given an undirected complete Euclidean weighted graph $$G(V, E)$$ and its MST $$T$$. I need to remove an arbitrary vertex $$v_i \in V(G)$$, and given a vertex $$v_j \notin V(G)$$, I have to calculate the MST of $$G^{‘}((V(G) \backslash \{v_i\})\cup\{v_j\}, (E\backslash\{(v_i, v_k): v_k \in V(G)\})\cup\{(v_k, v_j): v_k \in V(G^{‘})\})$$, i.e, the graph $$G$$ with the vertex $$v_j$$ (and its respective edges) and without the vertex $$v_i$$ (and its respective edges). To solve this, we can apply some well-know MST algorithms, such as Prim’s, Kruskal’s, Borukva’s algorithm. Neverthless, if we do this we would not use the already existing MST $$T$$, i.e., we would calculate a new whole MST. So, I would like to know if there is any way to reuse the existing MST $$T$$.

There are two a similar question to this here (with edges, considering only the removing of them), and here (with vertex, considering only the adding of them).