This problem asks for the longest possibly non-simple path through a tree, where some edges may be traversed at most twice and the rest at most once.

For the first subtask, we can write a brute-force backtracking solution to check all possible paths.

Time Complexity: $\mathcal{O}(2^N)$

For the second subtask, we can use a dynamic programming solution.

Root the tree at some node $r$. We will find two values for each node $u$:

1. $dp1[u]$: the length of the longest path starting and ending at $u$, completely contained within the subtree rooted at $u$.
2. $dp2[u]$: the length of the longest path starting in $r$ and ending in $u$.

To find $dp1[u]$, consider all children $v$ of $u$. If the edge $e$ (of length $d$) connecting $u$ and $v$ may be used twice, then we can go down to $v$, traverse $dp1[v]$, then go back up to $u$. Thus, $dp1[u]=\sum \{\begin{array}{ll}0& twice(e)=0\\ dp1[v]+2d& twice(e)=1\end{array}$.

To find $dp2[u]$, consider the parent $p$ of $u$. We can traverse $dp2[p]$, then traverse the $e$ connecting $p$ and $u$, then traverse $dp1[u]$. However, if $e$ may be used twice, $dp1[u]+2d$ would already have been counted under $dp2[p]$, so we must subtract off this value. Thus, $dp2[u]=\{\begin{array}{ll}dp2[p]+d+dp1[u]& twice(e)=0\\ dp2[p]-d& twice(e)=1\end{array}$.

The answer is then the maximum value over all $dp2[u]$ for all choices of $r$, for a total complexity of $\mathcal{O}(N^2)$.

Time Complexity: $\mathcal{O}(N^2)$

For the final subtask, we can expand on the solution for subtask 2.

Root the tree arbitrarily. Let $dp[u][c]$ be the length of the longest path that runs through node $u$ and is completely contained within the subtree rooted at $u$. There are 3 possible cases $c$:

1. $c=0$: The path starts and ends in $u$.
2. $c=1$: The path starts in $u$ and ends in some subtree of $u$.
3. $c=2$: The path starts and ends in two (not necessarily distinct) subtrees of $u$.

Let $e$ (of length $d$) be the edge connecting a node $u$ and its child $v$. Let's consider the largest possible lengths of 3 cases of paths going through both $u$ and $v$:

1. Loop (starts and ends in $u$): As described in the solution for subtask 2, if $e$ may be used twice, we can start at $u$, traverse $e$, traverse $dp[v][0]$, then traverse $e$ again to return to $u$. The length $le{n}_L(v)$ of the longest loop through $v$ would be $\{\begin{array}{ll}0& twice(e)=0\\ dp[v][0]+2d& twice(e)=1\end{array}$. Note that it is always optimal to loop through a child $v$ if possible; if you consider any path through $u$ that does not loop through $v$, adding such a loop will never decrease the path's length and will not affect the rest of the path.
2. 1-loose-end (starts in $u$ and ends in subtree $v$): We can start at $u$, traverse $e$, then traverse either $dp[v][0]$ or $dp[v][1]$. The length $le{n}_1(v)$ of the longest 1-loose-end through $v$ would be $max(dp[v][0],dp[v][1])+d$.
3. 2-loose-end (starts and ends in subtree $v$): If $e$ may be traversed twice, we can choose any of $dp[v][0]$, $dp[v][1]$, or $dp[v][2]$, then add the path $v\to u\to v$. The length $le{n}_2(v)$ of the longest 2-loose-end through $v$ would be $\{\begin{array}{ll}0& twice(e)=0\\ max(dp[v][0],dp[v][1],dp[v][2])+2d& twice(e)=1\end{array}$.

To find $dp[u][0]$, notice that such a path must consist entirely of loops. Thus, $dp[u][0]=\sum le{n}_L(v)$ over all children $v$ of $u$.

To find $dp[u][1]$, we can traverse the same path as $dp[u][0]$, but must replace one of the loops with a 1-loose-end. Thus, to maximize $dp[u][1]$, we must find the child of $u$, $v_{max}$, that has the greatest difference $le{n}_1(v_{max})-le{n}_L(v_{max})$. Therefore, $dp[u][1]=dp[u][0]-le{n}_L(v_{max})+le{n}_1(v_{max})$.

To find $dp[u][2]$, we can traverse the same path as $dp[u][0]$, but must do one of the following:

1. Replace two of the loops with two 1-loose-ends. We must find the two distinct children of $u$, $v_{max1}$ and $v_{max2}$, that have the greatest differences $le{n}_1(v_{max1})-le{n}_L(v_{max1})$ and $le{n}_1(v_{max2})-le{n}_L(v_{max2})$. Thus, the length of the longest such path is $dp[u][0]-le{n}_L(v_{max1})+le{n}_1(v_{max1})-le{n}_L(v_{max2})+le{n}_1(v_{max2})$.
2. Replace one of the loops with a 2-loose-end. We must find the child $v_{max3}$ that has the greatest difference $le{n}_2(v_{max3})-le{n}_L(v_{max3})$. Thus, the length of the longest such path is $dp[u][0]-le{n}_L(v_{max3})+le{n}_2(v_{max3})$.

Therefore, $dp[u][2]=max\{\begin{array}{l}dp[u][0]-le{n}_L(v_{max1})+le{n}_1(v_{max1})-le{n}_L(v_{max2})+le{n}_1(v_{max2})\\ dp[u][0]-le{n}_L(v_{max3})+le{n}_2(v_{max3})\end{array}$.

The final answer is then the maximum value over all $dp[u][c]$.

Time Complexity: $\mathcal{O}(N)$