1373. 二叉搜索子树的最大键值和

# 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 __init__(self):self.maxSum = 0def maxSumBST(self, root: Optional[TreeNode]) -> int:self.findMaxMinSum(root)return self.maxSumdef findMaxMinSum(self, root: TreeNode):""":param root::return:res[0]：root为根的二叉树是否为BST，若为1，则是BST；若为0，则不是BSTres[1]: 记录以root为根的二叉树所有节点的最小值res[2]: 记录以root为根的二叉树所有节点的最大值res[1]: 记录以root为根的二叉树所有节点值之和"""# base caseif not root:return [1, float('inf'), float('-inf'), 0]# 递归计算左右子树left = self.findMaxMinSum(root.left)right = self.findMaxMinSum(root.right)res = [None] * 4if left[0] == 1 and left[2] < root.val and right[0] == 1 and right[1] > root.val:res[0]= 1# 更新以root为根的这棵BST的最小值res[1] = min(left[1], root.val)# 更新以root为根的这棵BST的最大值res[2] = max(right[2], root.val)res[3] = left[3] + right[3] + root.val# 更新全局变量self.maxSum = max(self.maxSum, res[3])else:res[0] = 0#     其他值没必要计算了，用不到return res