graph - Time Complexity of modified dfs algorithm -

i want write algorithm finds optimal vertex cover of tree in linear time o(n), n number of vertices of tree.

a vertex cover of graph g=(v,e) subset w of v such every edge (a,b) in e, in w or b in w.

in vertex cover need have @ least 1 vertex each edge.

if pick non-leaf, can cover more 1 edge.

that's why thought can follows:

we visit root of tree, visit 1 of children, child of latter visited , on.

then if have reached @ leaf, check if have taken father optimal vertex cover, , if not pick it. then, if vertex picked optimal vertex cover has other children, pick first of them , visit leftmost children recursively , if reach @ leaf , father hasn't been chosen desired vertex cover, choose , on.

i have written following algorithm:

   dfs(node x){       discovered[x]=1;       each (v in adj(x)){           if discovered[v]==0{              dfs(v);              if (v->taken==0){                  x<-taken=1;              }           }       }     } 

i thought time complexity

(|v_i|, |e_i| number of vertices , edges respectively of subtrees @ root of call dfs )

is time complexity found right? or have calculated wrong?

edit: complexity of algorithm described recurrence relation:

t(|v|)=e*t(|v|-1)+o(1)

? or wrong?