LeetCode 938 Range Sum of BST

LeetCode

<div>

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:

<pre>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. </pre>

Example 2:

<pre>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. </pre>

Constraints:

• The number of nodes in the tree is in the range [1, 2 * 10<sup>4</sup>].
• 1 <= Node.val <= 10<sup>5</sup>
• 1 <= low <= high <= 10<sup>5</sup>
• All Node.val are unique.

</div>

题目大意：

给定[low, high]和BST，求满足条件的BST的节点和

解题思路：

Easy题，DFS，条件比较容易错

解题步骤：

N/A

注意事项：

1. 两个条件，若root.val在范围内，加入和。若low小于root.val(这里不取等号，因为所有节点是唯一，不存在相等节点)， 表示范围适用于左节点，同理右节点。

Python代码：

def rangeSumBST(self, root: TreeNode, low: int, high: int) -> int:
	if not root:
		return 0
	res = 0
	if low <= root.val <= high:
		res += root.val
	if low < root.val:
		res += self.rangeSumBST(root.left, low, high)
	if root.val < high:
		res += self.rangeSumBST(root.right, low, high)
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(1)