C++ SOLUTION BEATS 100% DETAILED COMMENTS
\n# Code\n\nclass Solution {\npublic:\n    // Function to find the maximum value sum\n    long long maximumValueSum(vector<int>& values, int k, vector<vector<in
\n# Code\n```\nclass Solution {\npublic:\n    // Function to find the maximum value sum\n    long long maximumValueSum(vector<int>& values, int k, vector<vector<int>>& edges) {\n        int n = values.size(); // Number of nodes\n        // Creating an adjacency list to represent the graph\n        vector<int> adjacencyList[n];\n\n        // Building the adjacency list\n        for(int i = 0; i < n - 1; i++) {\n            adjacencyList[edges[i][0]].push_back(edges[i][1]);\n            adjacencyList[edges[i][1]].push_back(edges[i][0]);\n        }\n\n        vector<int> nodesWithIncreasedValue; // Store the nodes whose value can be increased\n        // Loop through each node\'s value and determine if it can be increased by XOR operation with k\n        for(int i = 0; i < n; i++) {\n            long long newValue = (long long)values[i] ^ k; // Calculate the new value using XOR\n            if(newValue > values[i]) // If the new value is greater than the current value\n                nodesWithIncreasedValue.push_back(i); // Store the index of the node\n        }\n\n        int increasedNodesCount = nodesWithIncreasedValue.size(); // Count of nodes with increased value\n\n        if(increasedNodesCount % 2 == 0) { // If the count of increased nodes is even\n            long long totalSum = 0; // Initialize the total sum of values\n            for(int i = 0; i < n; i++) {\n                totalSum += values[i]; // Calculate the total sum of values\n            }\n            // Adjust the total sum by adding the increased values and subtracting the original values\n            for(int i = 0; i < increasedNodesCount; i++) {\n                totalSum -= values[nodesWithIncreasedValue[i]]; // Subtract the original value\n                long long newValue = (long long)values[nodesWithIncreasedValue[i]] ^ k; // Calculate the new value\n                totalSum += newValue; // Add the new value\n            }\n            return totalSum; // Return the final total sum\n        } else { // If the count of increased nodes is odd\n            long long totalSum = 0; // Initialize the total sum of values\n            for(int i = 0; i < n; i++) {\n                totalSum += (long long)values[i]; // Calculate the total sum of values\n            }\n            long long finalTotalSum = totalSum; // Initialize the final total sum with the original total sum\n\n            // Adjust the total sum by adding the increased values and subtracting the original values\n            for(int i = 0; i < increasedNodesCount; i++) {\n                totalSum -= values[nodesWithIncreasedValue[i]]; // Subtract the original value\n                long long newValue = (long long)values[nodesWithIncreasedValue[i]] ^ k; // Calculate the new value\n                totalSum += newValue; // Add the new value\n            }\n\n            // Try different combinations of nodes to maximize the total sum\n            for(int i = 0; i < increasedNodesCount; i++) {\n                long long newValue = (long long)values[nodesWithIncreasedValue[i]] ^ k; // Calculate the new value\n                long long currentTotalSum = totalSum - newValue + values[nodesWithIncreasedValue[i]]; // Adjust the total sum\n                finalTotalSum = max(finalTotalSum, currentTotalSum); // Update the final total sum if necessary\n            }\n\n            // Explore other nodes to maximize the total sum\n            vector<int> visitedNodes(n, 0); // Initialize visited nodes\n            for(int i = 0; i < increasedNodesCount; i++) {\n                visitedNodes[nodesWithIncreasedValue[i]] = 1; // Mark the increased nodes as visited\n            }\n\n            // Iterate through all nodes to find optimal combinations\n            for(int i = 0; i < n; i++) {\n                if(visitedNodes[i] == 0) { // If the node is not visited\n                    long long newValue = (long long)values[i] ^ k; // Calculate the new value\n                    long long currentTotalSum = totalSum - values[i] + newValue; // Adjust the total sum\n                    finalTotalSum = max(finalTotalSum, currentTotalSum); // Update the final total sum if necessary\n                }\n            }\n            return finalTotalSum; // Return the final total sum\n        }\n    }\n};\n\n\n\n```
['C++']