给定二叉搜索树的根结点 root，返回值位于范围 [low, high] 之间的所有结点的值的和。

示例 1：

输入：root = [10,5,15,3,7,null,18], low = 7, high = 15
输出：32

示例 2：

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

提示：

• 树中节点数目在范围 [1, 2 * 104] 内
• 1 <= Node.val <= 105
• 1 <= low <= high <= 105
• 所有 Node.val 互不相同

问题简要描述：返回位于范围 [low, high] 之间的所有结点的值的和 

Java 

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public int rangeSumBST(TreeNode root, int low, int high) {return dfs(root, low, high);}int dfs(TreeNode root, int low, int high) {if (root == null) {return 0;}int x = root.val;int ans = low <= x && x <= high ? x : 0;if (x > low) {ans += dfs(root.left, low, high);}if (x < high) {ans += dfs(root.right, low, high);}return ans;}
}

 Python3

# 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(root):if root is None:return 0x = root.valans = x if low <= x and x <= high else 0if x > low:ans += dfs(root.left)if x < high:ans += dfs(root.right)return ansreturn dfs(root)        

TypeScript

/*** Definition for a binary tree node.* class TreeNode {*     val: number*     left: TreeNode | null*     right: TreeNode | null*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {*         this.val = (val===undefined ? 0 : val)*         this.left = (left===undefined ? null : left)*         this.right = (right===undefined ? null : right)*     }* }*/function rangeSumBST(root: TreeNode | null, low: number, high: number): number {const dfs = (root) => {if (!root) {return 0;}let x = root.val;let ans = low <= x && x <= high ? x : 0;if (x > low) {ans += dfs(root.left);}if (x < high) {ans += dfs(root.right);}return ans;}return dfs(root);    
};