思路： 明確二叉搜尋樹的性質。按中序遍歷後得到的陣列是升序排列的。
L和R中間的數其實也相當於[L, R]之間的數。這樣有幾種方法可以得到：

注意： root = [10, 5, 15, 3, 7, 13, 18, 1, null, 6] 是按層從左到右給出的。

方法一、直接遍歷二叉樹，遍歷過程中直接把[L, R]之間的數相加。

遍歷條件：	if node.val < R:
                    stack.append(node.right) 
                if node.val > L:
                    stack.append(node.left)
 

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def rangeSumBST(self, root, L, R):
        """
        :type root: TreeNode
        :type L: int
        :type R: int
        :rtype: int
        """
        
        res, stack = 0, [root]
        while stack:
            node = stack.pop()
            if node:    # 只考慮不為空的節點
                if L<=node.val<=R:
                    res += node.val
                if node.val < R:
                    stack.append(node.right) 
                if node.val > L:
                    stack.append(node.left)
        return res
 

方法二、中序遍歷後得到有序陣列，再次遍歷得到L和R的位置，將位置間的數字想加。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def rangeSumBST(self, root, L, R):
        """
        :type root: TreeNode
        :type L: int
        :type R: int
        :rtype: int
        """
        global result
        result = []
        res = self.inorder(root)
        left = right = 0
        for i, val in enumerate(res):
            if val == L:
                left = i
            elif val == R:
                right = i
                break
        return sum(res[left:right+1])
    
    def inorder(self, root):
        if root is None:
            return
        self.inorder(root.left)
        result.append(root.val)
        self.inorder(root.right)
        return result
 

中間參考了：https://blog.csdn.net/qq_17550379/article/details/84034131