Make array P, where P[u] is sum of elements on path 1->u. Now problem is to check if there are u, v such that P[u] — P[v] = k. To solve it use dfs and store cnt structure.

When you are visiting u, cnt[x] will be the count of such v that P[v] = x and v belongs 1->u. If there is such u that cnt[P[u]-k] isn't zero, tree is good. All you need is to modify cnt to make it valid.

There is some code, i didn't check it: link