https://www.acmicpc.net/problem/1967
The definition of the diameter is simple.
The diameter of a tree is the maximum distance between one node to another node.
We say that the diameter of a circle is the longest distance between one point to another in a circle, so two ideas are kind of similar.
I tried several approaches and I found there is a simple method to find it.
First, these are the solutions I tried.
 Picking the maximum value of all distances between one node to another
This is the simplest way and with DFS, we can just get all distances between two arbitrary nodes, $i$ and $j$. But it takes long as the time complexity is bounded by $O(n^2)$.
 Extracting all leaf nodes and finding maximum distances from one leaf to other
To become the diameter, two nodes should be separated as far as possible, so it is obvious that these two nodes are both leaf nodes according to the structure of a tree. So I filtered leaves only and executed DFS, but the running time still exceeded. The time complexity is still $O(n^2)$.
 Based on method 2, making the function not search outside of the common parent node of two leaf nodes
This can reduce the time a little bit, but it does not change $O(n^2)$ time complexity. And finding a common parent is also quite challenging work.
In conclusion, this is an optimal solution.
Finding a tree’s diameter consists of three steps.
 Select an arbitrary node $x$ and find the node $y$ which is the farthest node from $x$.
 Find the node $t$ which is located in the farthest spot from the location of $y$.
 Calculate the distance between $y$ and $t$.
It turns out that step 2 and 3 can be conducted at the same time.
In other words, we can get an answer by executing DFS two times.
So the time complexity becomes $O(n)$.
It is not hard to apply the above algorithm, but the problem is why that method gets the diameter correctly.
This is how to prove it.
If there are node $u$ and $v$ and the distance between them is the diameter, then after picking one arbitrary node $x$ and finding the farthest node $y$, $y$ becomes $u$ or $v$.
1. The case that x is $u$ or $v$
This case obviously proves above statement because the farthest node from $x$, which is $y$, would be $u$ or $v$.
2. The case that y is $u$ or $v$
Same as 1.
3. The case that $x$ and y* are different from $u$ and $v$
This case can be divided into 2 specific cases.

The case that $x$$y$ does not meet $u$$v$ As we can see from the image, we can think of the path $t$$r$ by picking one point from each path. By making this kind of new path, $d(x, y)$ becomes bigger than $d(x, v)$. In other words, it is a contradiction to the assumption that $y$ is the farthest node from $x$.

The case that $x$$y$ meets $u$$v$
In this case, two paths share at least one node. Let’s say that they share only one spot. Because $d(x, y)$ is the distance between $x$ and the farthest node $y$, $d(x, y)>d(x, v)$. It leads to $d(t, y)>d(t, v)$. Then $d(u, y) = d(u, t) + d(t, y) >d(u, t) + d(t, v) = d(u, v)$, so there is a longer path than $u$$v$, which is a contradiction to the assumption that $u$$v$ is the diameter. So this case cannot exist and the same when there are multiple points shared.
In conclusion, case 3 cannot be possible so there are only two cases 1 and 2.
So the above algorithm can get the correct diameter.
I built the codes assuming that the arbitrary node $x$ is the root node.
Here is the final example.
#include <iostream>
#include <vector>
#include <stack>
using namespace std;
int n;
vector<pair<int, int>> graph[10001];
pair<int, int> findMax(int start) {
bool checked[n+1] = {false, };
stack<pair<int, int>> s;
s.push({start, 0});
checked[start] = true;
pair<int, int> result (1, 0);
while(1) {
if (s.empty()) {
break;
}
pair<int, int> cur = s.top();
s.pop();
if (cur.second >= result.second) {
result.second = cur.second;
result.first = cur.first;
}
for (int i=0; i<graph[cur.first].size(); ++i) {
if (!checked[graph[cur.first][i].first]) {
s.push({graph[cur.first][i].first, cur.second + graph[cur.first][i].second});
checked[graph[cur.first][i].first] = true;
}
}
}
return result;
}
int main() {
scanf("%d", &n);
for (int i=0; i<n1; ++i) {
int a, b, w;
scanf("%d %d %d", &a, &b, &w);
graph[a].push_back({b, w});
graph[b].push_back({a, w});
}
int max_node = findMax(1).first;
int max_len = findMax(max_node).second;
printf("%d\n", max_len);
return 0;
}