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:

    10
   /  \
  5    15
 / \    \
3   7    18

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:

       10
     /    \
    5      15
   / \    /  \
  3   7  13   18
 /   /
1   6

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.

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rangeSumBST(self, root: Optional[TreeNode], low: int, high: int) -> int:
        def dfs(node: Optional[TreeNode]):
            if not node:
                return
            if low <= node.val <= high:
                self.sum += node.val
            if node.val > low:
                dfs(node.left)
            if node.val < high:
                dfs(node.right)

        self.sum = 0
        dfs(root)
        return self.sum