C++ 🧠 Solution Explanation ✅
Intuition\nEvery node of the first tree must be linked to a node with the maximum number of target nodes for k = k - 1 in the second tree. \n\nFor instance, in 
# Intuition\nEvery node of the first tree must be linked to a node with the maximum number of target nodes for k = k - 1 in the second tree. \n\nFor instance, in the first example, such a node in the second tree is either node `0` or node `4` with maximum number of 4 target nodes. \n\nWe find the number of nodes target to node `i` in the first tree and add the maximum value (for exmaple 1, it is 4). \n\nWe use `DFS` to find the number of nodes target to a node `i` in a tree.\n\n\n\n# Complexity\n- Time complexity:\n`O(n^2)` where n is the number of nodes in the first tree. \n\n- Space complexity:\n`O(1)` \n\n# Code\n```cpp []\n#define vv vector\n\n\nint dfs(vv<vv<int>> &tree, int v, int p, int k)\n{\n    if (k == 0)\n        return 1;\n\n    int ans = 1;\n    for (int u : tree[v])\n    {\n        if (u != p)\n        {\n            ans += dfs(tree, u, v, k - 1);\n        }\n    }\n\n    return ans;\n}\n\nclass Solution {\npublic:\n    vector<int> maxTargetNodes(vector<vector<int>>& edges1, vector<vector<int>>& edges2, int k) {\n        int n = edges1.size() + 1;\n    int m = edges2.size() + 1;\n    vv<vv<int>> tree1(n), tree2(m);\n    for (auto &edge : edges1)\n    {\n        tree1[edge[0]].push_back(edge[1]);\n        tree1[edge[1]].push_back(edge[0]);\n    }\n\n    for (auto &edge : edges2)\n    {\n        tree2[edge[0]].push_back(edge[1]);\n        tree2[edge[1]].push_back(edge[0]);\n    }\n\n    int mx = 0;\n    for (int i = 0; i < m && k != 0; i++)\n    {\n        mx = max(mx, dfs(tree2, i, -1, k - 1));\n    }\n\n    vv<int> ans(n);\n    for (int i = 0; i < n; i++)\n    {\n        ans[i] = dfs(tree1, i, -1, k) + mx;\n    }\n\n    return ans;\n    }\n};\n```
['C++']