C++ || Thinking Process And Explanation
Intuition\n1. Consider the property of the xor operation, where any number do xor twice remains unchanged, so considering how can we apply this property to get 
# Intuition\n1. Consider the property of the `xor` operation, where any number do `xor` twice remains unchanged, so considering how can we apply this property to get the result.\n2. Given the `edges`, we can perform the `xor` operation on two adjacent nodes. However, it seems not using the property, we think beyond adjacent nodes and consider applying the `xor` operation to any two nodes in the tree.\n3. Since the input is a tree, any two nodes are connected by a path. When performing `xor` on the values of two nodes, all intermediate nodes on the path (excluding the start and end nodes) will undergo the `xor` operation twice, remaining unchanged. Thus, we can perform the `xor` operation on any two nodes.\n4. As a result, The problem reduces to selecting any two nodes at a time, performing the `xor` operation, and maximizing the sum of the resulting values.\n5. Calculate the potential change in value (delta) for each node if `xor` with `k`, and store these deltas in a vector. Sort this delta vector in descending order.\n6. Traverse the delta vector, select two delta, and add them to the final result in each round if the sum of them contributes positively to the final answer.\n\n# Approach\n1. Calculate the potential delta for each node value if `xor` with `k`.\n2. Sort the delta values in descending order.\n3. Traverse the delta vector and add 2 delta values to final result at a time in each round if they contribute the final answer.\n\n# Complexity\n- Time complexity: $$O(n*log(n))$$\n- Space complexity: $$O(n)$$\n\n# Code\n```\nclass Solution {\npublic:\n    long long maximumValueSum(vector<int>& nums, int k, vector<vector<int>>& edges) {\n        int n = nums.size();\n        vector<int> delta;\n        for (int i = 0; i < n; i++) {\n            int value = nums[i];\n            delta.push_back((value^k)-value);\n        }\n        sort(delta.rbegin(), delta.rend());\n\n        long long res = accumulate(begin(nums), end(nums), 0LL);\n        for (int i = 0; i+1 < n; i+= 2) {\n            int delta_1 = delta[i];\n            int delta_2 = delta[i+1];\n            if (delta_1+delta_2 > 0) {\n                res += delta_1+delta_2;\n            }\n        }\n        return res;\n    }\n};\n\n```
['C++']