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Author: assassint2017

Range Sum of BST/Range_Sum_of_BST.cpp

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+// 使用二叉树中序遍历的解法
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+class Solution 
+{
+public:
+    int rangeSumBST(TreeNode* root, int L, int R) 
+    {
+        stack<TreeNode*> memo;
+        int res = 0;
+        while (root || !memo.empty())
+        {
+            while (root)
+            {
+                memo.push(root);
+                root = root->left;
+            }
+            if (!memo.empty())
+            {
+                root = memo.top();
+                memo.pop();
+                
+                if (root->val >= L && root->val <= R)
+                    res += root->val;
+                
+                root = root->right;
+            }
+        }
+        
+        return res;
+    }
+};
+
+
+// 第二种做法 使用递归
+// Runtime: 144 ms, faster than 84.60% of C++ online submissions for Range Sum of BST.
+// Memory Usage: 41.2 MB, less than 80.00% of C++ online submissions for Range Sum of BST.
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+class Solution 
+{
+public:
+    int rangeSumBST(TreeNode* root, int L, int R) 
+    {
+        if (root == nullptr || L > R)
+            return 0;
+        else if (L == R && L == root->val)
+            return root->val;
+        else if (L > root->val)
+            return rangeSumBST(root->right, L, R);
+        else if (root->val > R)
+            return rangeSumBST(root->left, L, R);
+        else
+            return rangeSumBST(root->left, L, root->val - 1) + rangeSumBST(root->right, root->val + 1, R) + root->val;
+    }
+};