I guess the solution would look something like this:

Let $$$n$$$ be the number of nodes, let $$$i = 0$$$ be the root node and let $$$C_i$$$ be the set of all direct children of $$$i$$$ for any node $$$i$$$. We will define a dp array $$$dp[n][3]$$$ as follows:

States:

• $$$dp[i][0]$$$ equals the minimum number of nodes in $$$S$$$ in the subtree of $$$i$$$ such that $$$i \in S$$$ and all nodes in the subtree of $$$i$$$ (not including $$$i$$$ itself) are valid according to the statement.
• $$$dp[i][1]$$$ equals the minimum number of nodes in $$$S$$$ in the subtree of $$$i$$$ such that $$$i \not \in S$$$ but $$$i$$$ has a child in $$$S$$$ and all nodes in the subtree of $$$i$$$ (not including $$$i$$$ itself) are valid according to the statement.
• $$$dp[i][2]$$$ equals the minimum number of nodes in $$$S$$$ in the subtree of $$$i$$$ such that $$$i \not \in S$$$ and $$$i$$$ doesn't have a child in $$$S$$$ and all nodes in the subtree of $$$i$$$ (not including $$$i$$$ itself) are valid according to the statement.

For all $$$dp[i][j]$$$ if there exists no solution, we will have $$$dp[i][j] > n$$$ to denote that no solution exists. This works because for a valid solution, the number of nodes in $$$S$$$ can never exceed $$$n$$$. This is done in order to make implementation easier.

Transitions:

If $$$|C_i| = 0$$$ ($$$i$$$ has no children):

• $$$dp[i][0] = 1$$$ (we can add $$$i$$$ to $$$S$$$; this is valid and $$$|S| = 1$$$)
• $$$dp[i][1] = 10^6$$$ (there is no way for $$$i$$$ to have a child in $$$S$$$ as $$$i$$$ has no children. Also note that any value $$$> n$$$ works here, not just $$$10^6$$$. Here I'm assuming that $$$n \le 2 \cdot 10^5$$$)
• $$$dp[i][2] = 0$$$ (we can choose to not add $$$i$$$ to $$$S$$$; this is valid (for now) and $$$|S| = 0$$$)

If $$$|C_i| > 0$$$:

• $$$dp[i][0] = \displaystyle\sum_{j \in C_i}dp[j][2]$$$ (in order to have $$$i \in S$$$, no children of $$$i$$$ can be in $$$S$$$ and also no children of $$$i$$$ can have their children in $$$S$$$ so that the children of $$$i$$$ are not adjacent to multiple nodes in $$$S$$$)
• $$$dp[i][1] = \displaystyle\min_{k \in C_i}(dp[k][0]+\sum_{j \in C_i \setminus k}dp[j][1])$$$ (in order to have $$$i \not \in S$$$ but a child of $$$i$$$ in $$$S$$$, exactly one child of $$$i$$$ has to be in $$$S$$$. As $$$i \not \in S$$$, all other children of $$$i$$$ have to have a child in $$$S$$$ in order to be adjacent to some node in $$$S$$$)
• $$$dp[i][2] = \displaystyle\sum{j \in C_i}dp[j][1]$$$ (in order to have $$$i \not \in S$$$ and no child of $$$i$$$ in $$$S$$$, all children of $$$i$$$ have to have a child in $$$S$$$ in order to be adjacent to some node in $$$S$$$)

After the dp has been calculated up to the root, we will have three dp values for the root. $$$dp[0][2]$$$ is not a valid solution, as (by definition) $$$0 \not \in S$$$ and $$$0$$$ doesn't have any children in $$$S$$$. $$$dp[0][0]$$$ and $$$dp[0][1]$$$ are valid solutions and thus, the answer to the problem is $$$\min(dp[0][0], dp[0][1])$$$. If this answer is $$$> n$$$, no solution exists.

All transitions can be calculated in $$$O(|C_i|)$$$ and thus, the total complexity is $$$O(\sum |C_i|) = O(n)$$$. (The transition for $$$dp[i][1]$$$ can be calculated in $$$O(|C_i|)$$$ by first calculating $$$\displaystyle\sum_{j \in C_i}dp[j][1]$$$ in $$$O(|C_i|)$$$ and then calculating the value for each $$$k \in C_i$$$ as $$$dp[k][0]+\left(\displaystyle\sum_{j \in C_i}dp[j][1]\right) - dp[k][1]$$$ in $$$O(|C_i|)$$$ and then taking the minimum.)

Please let me know if you think there is something wrong with my approach or if I've made a mistake somewhere.