Let the graph Ln be defind by VLnv1v2VnELnV1v2vn1vn show tha

Let the graph L_n be defind by: V(L_n)={v_1,v_2,...V_n},E(L_n)={(V_1,v_2),...(v_n-1,v_n)} show that if a tree width n verticals has an Euler walk, it must be isomorphic to L_n

Solution

if there exists a eulerian walk in a tree , trace the tree along the walk , since tree cannot have a cycle the walk is actually a path. So there exists a path which goes to every vertex ( since the path includes all the edges of tree , it includes all the vertices from the tree) . Hence the tree is simply L_n