Easy and Fast Solution using DFS and tuple
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    vector<long long> placedCoins(vector<vector<int>>& edges, vector<int>& cost) {\n        long long n = cost.size();\n        \n        // Create the adjacency list for the tree\n        vector<vector<long long>> tree(n);\n        for (const auto& edge : edges) {\n            tree[edge[0]].push_back(edge[1]);\n            tree[edge[1]].push_back(edge[0]);\n        }\n        \n        vector<long long> coin(n, 0);\n        vector<long long> subtree_size(n, 0);\n\n        // Helper function to update the top three positive and negative values\n        auto updateTop3 = [](tuple<long long, long long, long long>& top3, long long value) {\n            auto& [max1, max2, max3] = top3;\n            if (value > max1) {\n                max3 = max2;\n                max2 = max1;\n                max1 = value;\n            } else if (value > max2) {\n                max3 = max2;\n                max2 = value;\n            } else if (value > max3) {\n                max3 = value;\n            }\n        };\n\n        auto updateBottom3 = [](tuple<long long, long long, long long>& bottom3, long long value) {\n            auto& [min1, min2, min3] = bottom3;\n            if (value < min1) {\n                min3 = min2;\n                min2 = min1;\n                min1 = value;\n            } else if (value < min2) {\n                min3 = min2;\n                min2 = value;\n            } else if (value < min3) {\n                min3 = value;\n            }\n        };\n\n        // DFS to calculate subtree sizes and coins\n        function<pair<tuple<long long, long long, long long>, tuple<long long, long long, long long>>(long long, long long)> dfs = [&](long long node, long long parent) {\n            subtree_size[node] = 1;\n            \n            // Track the top three greatest positive and negative values from children\n            tuple<long long, long long, long long> top3 = {LLONG_MIN, LLONG_MIN, LLONG_MIN};\n            tuple<long long, long long, long long> bottom3 = {LLONG_MAX, LLONG_MAX, LLONG_MAX};\n            \n            // Traverse the children\n            for (long long neighbor : tree[node]) {\n                if (neighbor == parent) continue;\n                \n                // Perform DFS on the child and get its top three positive and negative values\n                auto [child_top3, child_bottom3] = dfs(neighbor, node);\n\n                // Compare and update the top three values for the current node\n                updateTop3(top3, get<0>(child_top3));\n                updateTop3(top3, get<1>(child_top3));\n                updateTop3(top3, get<2>(child_top3));\n\n                // Compare and update the bottom three values for the current node\n                updateBottom3(bottom3, get<0>(child_bottom3));\n                updateBottom3(bottom3, get<1>(child_bottom3));\n                updateBottom3(bottom3, get<2>(child_bottom3));\n\n                // Update the subtree size\n                subtree_size[node] += subtree_size[neighbor];\n            }\n\n            // Include the current node\'s cost in both top and bottom tracking\n            updateTop3(top3, cost[node]);\n            updateBottom3(bottom3, cost[node]);\n\n            // If the size of the subtree is less than 3, place 1 coin\n            if (subtree_size[node] < 3) {\n                coin[node] = 1;\n            } else {\n                // Calculate the maximum product of the top 3 distinct positive costs\n                auto [max1, max2, max3] = top3;\n                auto [min1, min2, min3] = bottom3;\n\n                long long max_product = LLONG_MIN;\n                \n                // Option 1: Product of top 3 positive values\n                if (max3 != LLONG_MIN) {  // Ensure we have 3 valid distinct costs\n                    max_product = max(max_product, 1LL * max1 * max2 * max3);\n                }\n\n                // Option 2: Product of two most negative values and one positive value\n                if (min2 != LLONG_MAX && max1 != LLONG_MIN) {  // Ensure we have 2 valid negative and 1 valid positive cost\n                    max_product = max(max_product, 1LL * min1 * min2 * max1);\n                }\n\n                coin[node] = max(max_product, 0LL);  // If the product is negative or invalid, place 0 coins\n            }\n\n            // Return the top three positive and bottom three negative values found\n            return make_pair(top3, bottom3);\n        };\n        \n        // Start DFS from the root (node 0)\n        dfs(0, -1);\n        \n        return coin;\n    }\n};\n\n```
['C++']