# 938. Range Sum of BST

Given the `root` node of a binary search tree and two integers `low` and `high`, return the sum of values of all nodes with a value in the inclusive range `[low, high]`.

Example 1:

```Input: root = [10,5,15,3,7,null,18], low = 7, high = 15
Output: 32
Explanation: Nodes 7, 10, and 15 are in the range [7, 15]. 7 + 10 + 15 = 32.
```

Example 2:

```Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
Output: 23
Explanation: Nodes 6, 7, and 10 are in the range [6, 10]. 6 + 7 + 10 = 23.
```

Constraints:

• The number of nodes in the tree is in the range `[1, 2 * 104]`.
• `1 <= Node.val <= 105`
• `1 <= low <= high <= 105`
• All `Node.val` are unique.

Solution:

DFS recursive:

``````class Solution:
def rangeSumBST(self, root, L, R):
"""
:type root: TreeNode
:type L: int
:type R: int
:rtype: int
"""
def dfs(root):
if not root:
return
if L <= root.val <= R:
self.res += root.val
if L <= root.val:
dfs(root.left)
if R >= root.val:
dfs(root.right)
self.res = 0
dfs(root)
return self.res
``````

DFS iterative:

``````class Solution:
def rangeSumBST(self, root, L, R):
"""
:type root: TreeNode
:type L: int
:type R: int
:rtype: int
"""
stack = [root]
res = 0
while stack:
u = stack.pop()
if L <= u.val <= R:
res += u.val
if u.left and u.val >= L:
stack.append(u.left)
if u.right and u.val <= R:
stack.append(u.right)
return res``````
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