Create 3203. Find Minimum Diameter After Merging Two Trees

3203. Find Minimum Diameter After Merging Two Trees

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+class Solution {
+public:
+    void dfs(int node, int par, int dis, int &maxi, vector<vector<int>>&adj, int &far){
+        if(dis>maxi){
+            maxi = dis;
+            far = node;
+        }
+        for(auto it:adj[node]){
+            if(it!=par){
+                dfs(it, node, dis+1, maxi, adj, far);
+            }
+        }
+    }
+
+    int minimumDiameterAfterMerge(vector<vector<int>>& edges1, vector<vector<int>>& edges2) {
+        int n = edges1.size()+1;
+        int m = edges2.size()+1;
+        vector<vector<int>>adj1(n);
+        vector<vector<int>>adj2(m);
+        for(auto it: edges1){
+            adj1[it[0]].push_back(it[1]);
+            adj1[it[1]].push_back(it[0]);
+        }
+
+        int maxi1 = 0;
+        int far = 0;
+        dfs(0, -1, 0, maxi1, adj1, far);
+
+        maxi1 = 0;
+        int far1 = 0;
+        dfs(far, -1, 0, maxi1, adj1, far1);
+
+        for(auto it: edges2){
+            adj2[it[0]].push_back(it[1]);
+            adj2[it[1]].push_back(it[0]);
+        }
+
+        int maxi2 = 0;
+        far = 0;
+        dfs(0, -1, 0, maxi2, adj2, far);
+
+        maxi2 = 0;
+        int far2 = 0;
+        dfs(far, -1, 0, maxi2, adj2, far2);
+
+        int tmp1 = (maxi1+1)/2;
+        int tmp2 = (maxi2+1)/2;
+        int mrg = tmp1+tmp2+1;
+        return max(maxi1, max(mrg, maxi2));
+    }
+};