> For the complete documentation index, see [llms.txt](https://cai-sen-se.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://cai-sen-se.gitbook.io/leetcode/1001-2000/1373.-er-cha-sou-suo-zi-shu-de-zui-da-jian-zhi-he.md).

# 1373.二叉搜索子树的最大键值和

<https://leetcode-cn.com/problems/maximum-sum-bst-in-binary-tree/>

只要把前序遍历变成后序遍历，让 `traval` 函数把辅助函数做的事情顺便做掉

```python
class Solution:
    maxSum = 0

    def maxSumBST(self, root: TreeNode) -> int:
        #遍历函数
        def traval(root):
            if root == None:
                return [1, sys.maxsize, -sys.maxsize, 0]
            left = traval(root.left)
            right = traval(root.right)
            #res[0] 记录以 root 为根的二叉树是否是 BST，若为 1 则说明是 BST，若为 0 则说明不是 BST；
            #res[1] 记录以 root 为根的二叉树所有节点中的最小值；
            #res[2] 记录以 root 为根的二叉树所有节点中的最大值；
            #res[3] 记录以 root 为根的二叉树所有节点值之和。
            res = [0]*4
            #当左右子树都为BST，且左子都比root小，右子都比root大
            if left[0]==1 and right[0] == 1 and root.val > left[2] and root.val < right[1]:
                res[0]= 1
                res[1] = min(left[1], root.val)
                res[2] = max(right[2], root.val)
                res[3] = left[3] + right[3] + root.val
                self.maxSum = max(self.maxSum, res[3])
            else:
                res[0] = 0
            return res
				#开始遍历
        traval(root)
        return self.maxSum
```


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