Consider any node as the root of the tree.

Let A[i]= minimum number of tickets for subtree rooted at node i and B[i]= minimum number of tickets for subtree rooted at node i where node i can be connected to at most one node.

Note that B[i]>=A[i].

for every node i, use dp to calculate A[j] and B[j] for all children j of i. Now We can calculate A[i] and B[i] as follows: There can be at most 3 cases.

1. node i is not connected to any node, so A[i]= 1+ sum(A[j]);
2. node i is connected to one node. Let x be the child of i whose B[j]-A[j] is minimum. Then we connect this child with i. So A[i]=sum(a[j])-A[x]+B[x];
3. node i is connected to two nodes. Then take two child x and y whose B[j]-A[j] is minimum. It is like the previous one . A[i]= sum(a[j])-A[x]+B[x]-A[y]+B[y]-1; Just take the minimum of this three results.

B[i] can be computed in similar way.