# 938.Range Sum of BST

**938.Range Sum of BST**

难度:Easy

> 给定二叉搜索树的根结点 root，返回 L 和 R（含）之间的所有结点的值的和。

二叉搜索树保证具有唯一的值。

```
示例 1：

输入：root = [10,5,15,3,7,null,18], L = 7, R = 15
输出：32
示例 2：

输入：root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
输出：23


提示：

树中的结点数量最多为 10000 个。
最终的答案保证小于 2^31。
```

方法:一般都是递归求解:

```
/**
 * 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) return 0;
        if(root->val >= L && root->val <=R)
            return root->val + rangeSumBST(root->left,L,R) + rangeSumBST(root->right,L,R);
        else
        return rangeSumBST(root->left,L,R) + rangeSumBST(root->right,L,R);
    }
};
```


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