MOST EASY PREORDER SOLUTION || EASY TO UNDERSTAND || C++
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```\n/**\n * Definition for a binary tree node.\n * struct TreeNode {\n *     int val;\n *     TreeNode *left;\n *     TreeNode *right;\n *     TreeNode() : val(0), left(nullptr), right(nullptr) {}\n *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}\n * };\n */\nclass Solution {\npublic:\n    int sum;\n    void rec(TreeNode* root, int low, int high){\n        if(root==NULL) return ;\n        if(root->val <= high && root-> val >= low){\n            sum += root->val;\n        }\n        rec(root->left , low,high);\n        rec(root->right , low , high);\n    }\n    int rangeSumBST(TreeNode* root, int low, int high) {\n        sum = 0;\n        rec(root, low,high);\n        return sum;\n    }\n};\n```
['C++']