---
# System prepended metadata

title: '[938. Range Sum of BST](https://leetcode.com/problems/range-sum-of-bst/)'
tags: [leetcode]

---

---
tags: leetcode
---

# [938. Range Sum of BST](https://leetcode.com/problems/range-sum-of-bst/)
---
# My Solution
## The Key Idea for Solving This Coding Question
## C++ Code 1
```cpp=
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

class Solution {
public:
    int rangeSumBST(TreeNode *root, int low, int high) {
        int answer = 0;

        dfs(root, low, high, answer);
        return answer;
    }

private:
    void dfs(TreeNode *root, int low, int high, int &answer) {
        if (root == nullptr) {
            return;
        }

        if (root->val < low) {
            dfs(root->right, low, high, answer);
            return;
        }

        if (high < root->val) {
            dfs(root->left, low, high, answer);
            return;
        }

        answer += root->val;
        dfs(root->left, low, high, answer);
        dfs(root->right, low, high, answer);
    }
};
```
### Time Complexity
$O(n)$
$n$ is the number of nodes in the binary tree referred by `root`.
### Space Complexity
$O(H)$
$H$ is the height of the binary tree referred by `root`.

## C++ Code 2
```cpp=
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int rangeSumBST(TreeNode* root, int low, int high) {
        if (!root) {
            return 0;
        }
        
        if (root->val < low) {
            return rangeSumBST(root->right, low, high);
        }
        
        if (high < root->val) {
            return rangeSumBST(root->left, low, high);
        }
        
        return root->val + 
               rangeSumBST(root->left, low, high) + 
               rangeSumBST(root->right, low, high);
    }
};
```
### Time Complexity
$O(n)$
$n$ is the number of nodes in the binary tree referred by `root`.
### Space Complexity
$O(H)$
$H$ is the height of the binary tree referred by `root`.
# Miscellane
<!--
# Test Cases
```
[10,5,15,3,7,null,18]
7
15
```
```
[10,5,15,3,7,13,18,1,null,6]
6
10
```
-->