Find a tree with degree sequence $(4,3,3,3,2,2,2,1,\ldots),$ where the number of vertices of degree $1$ is not specified and prove that any tree found must have the same number of vertices.
One possible tree is as follows. First consider the simple path $ABCDE.$ Then at the vertex $E,$ the tree will have three edges $EF, \ EG, \ EH.$ At the vertex $F,$ there are two edges $FI, \ FJ.$ Also, the vertex $G$ splits into $GK, \ GL.$ Finally, there are edges $HM, \ HN$ at the vertex $H.$
Assuming the tree above is correctly constructed and given $(4,3,3,3,2,2,2,1,\ldots),$ looks like the actual degree sequence must be $(4,3,3,3,2,2,2,1).$ Seems to me all the trees constructed from the sequence $(4,3,3,3,2,2,2,1)$ must have $8$ vertices.
Does the above make sense?