Given an undirected graph $G=(V,E)$, and vertex $v\in V$ and a number $k\in \mathbb{N}$, find an algorithm to find whether there exists a spanning tree of $G$ in which $v$ satisfies $d(v)=k$

I've narrowed the solution to 3 situations, in $G$: $d(v)<k , \\d(v)=k\\d(v)>k$

When $d(v)<k$, it is clear that there cannot exist a spanning tree of $G$ such that $d(v)=k$, since $d(v)$ was never of value $k$ to begin with.

When $d(v)>k$, we know that we can create a graph $G'$ in which we remove a finite amount of edges $i$ in such a matter that we reduce $d(v)$ to be $k$, and check whether $G'$ is connected or not (we can use BFS), since $d(v)\leq |V|$, the complexity would be $O(|V|\cdot (|V|+|E|))$, however I feel this is extremly inefficient.

For the case where $d(v)=k$ in $G$, I am uncertain on how to proceed.

EDIT: I've managed to update the case where $d(v)>k$ to be in linear time:

if $d(v)>k$, we'll have to remove edges from the graph.

We'll perform $DFS$ on $v$, since $d(v)>k$ there will be back edges, we can (in linear time) count how many back edges are going into(or out of) $v$, if the number of back edges is greater than $m-k$ where $|E|=m$, we can remove $m-k$ back edges from $G$, if the number of back edges is smaller than $m-k$, there cannot be a spanning tree in which $d(v)=k$.

The updated solution to the case was given to me by a colleague, I dont completely understand it yet I am trying to wrap my head around it.

About the case where $d(v)=k$, I am still uncertain.

  • 1
    $\begingroup$ Why can't you just run an MST algorithm in case $d(v) = k$? Or better yet if you do not need an MST, a BFS starting from $v$ to see if the graph is connected which would imply that there is a tree that spans all vertices, hence an MST algorithm will find one that minimizes the total weight. $\endgroup$
    – Russel
    Commented Jan 14, 2023 at 13:31
  • $\begingroup$ The question asks for a linear time complexity in all cases, I've edited my answer for the case in which $d(v)>k$ with a linear solution. MST finding algorithms run in $O(nlogn)$ $\endgroup$
    – Aishgadol
    Commented Jan 14, 2023 at 13:35
  • $\begingroup$ Please see my edited comment. And maybe include that running time requirement in the question $\endgroup$
    – Russel
    Commented Jan 14, 2023 at 13:36
  • $\begingroup$ @Russel Please note that this problem doesnt necessarily discuss MSTs, but spanning trees in general. $\endgroup$
    – Aishgadol
    Commented Jan 14, 2023 at 13:42
  • $\begingroup$ So I think running BFS starting from $v$ and checking if all vertices are reachable will suffice when the degree of $v$ in $G$ is $k$, since a BFS will force all edges of $v$ to be part of the BFS tree rooted at $v$, which will be a spanning tree if all vertices will be reached from $v$ $\endgroup$
    – Russel
    Commented Jan 14, 2023 at 13:45

1 Answer 1


First, verify that $G$ is indeed connected, otherwise say no.

Second, if $G - v$ (the graph after deleting $v$) has more than $k$ components, we can say no.

Pick any $k$ neighbors of $v$ with the restriction that you pick at least one from each component of $G-v$ and mark them. The edges from $v$ to these marked vertices are in the spanning tree.

For each connected component $C$ of $G-v$, pick a marked vertex $x$. Run a DFS from $x$ and pick the edges you traverse as long as you see an unvisited vertex. If you meet a visited vertex, you ignore the edge. If you meet a marked vertex, simply do not put that edge into the spanning tree.

  • $\begingroup$ The use of components here is a nice touch. What's the idea for the last part? Wouldn't the component be enough? $\endgroup$
    – Russel
    Commented Jan 15, 2023 at 2:36
  • $\begingroup$ My apologies, that was not very well explained. When you run DFS, simply do not add the edges to the marked vertices (except the starting point) to the spanning tree. $\endgroup$ Commented Jan 15, 2023 at 9:16

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