Use centroid decomposition) If your current centroid — v, and you dfs to u from v, there three case: 1) On path (v,u) there 2 nodes with value 1. Then you want chose any node, which is already dfsed from v, and on path between this node and v there 0 nodes with value 1. 2) On path (v,u) there 1 node with value 1. Then you want chose any node, which is already dfsed from v, and on path between this node and v there 1 node with value 1. 3) On path (v,u) there 0 nodes with value 1. Then you want chose any node, which is already dfsed from v, and on path between this node and v there 2 nodes with value 1.

And sorry for my english)

Root the tree at an arbitrary node. Define DP state like this —

dp[u][x] = how many x distance nodes (from u) with value 0 are there in subtree of u

By distance I mean number of 1's in the path. Also note that we don't need to know any answer for x > 2. You can calculate these values easily by a dfs.
So we are done with all chains hanging down. Only thing left is to cross chains.

To combine results, you can divide in two cases —

• Current node has value 1 — Then you can choose some 1 distance node from a subtree and 0 distance nodes from all other subtree. Or you can choose 0 distance node from a subtree and 1 distance node from all other subtree.

• Current node has value 0 — Then choose i-distance nodes from one subtree and 2 - i distance nodes from other subtree, .

You may need to adjust the final answer, depends on the way you implement it.