# leetcode\_1373

Given a binary tree root, the task is to return the maximum sum of all keys of any sub-tree which is also a Binary Search Tree (BST).

Assume a BST is defined as follows:

* The left subtree of a node contains only nodes with keys less than the node's key.
* The right subtree of a node contains only nodes with keys greater than the node's key.
* Both the left and right subtrees must also be binary search trees.

Example 1:

![](https://assets.leetcode.com/uploads/2020/01/30/sample_1_1709.png)

```
Input: root = [1,4,3,2,4,2,5,null,null,null,null,null,null,4,6]
Output: 20
Explanation: Maximum sum in a valid Binary search tree is obtained in root node with key equal to 3.
```

Example 2:

![](https://assets.leetcode.com/uploads/2020/01/30/sample_2_1709.png)

```
Input: root = [4,3,null,1,2]
Output: 2
Explanation: Maximum sum in a valid Binary search tree is obtained in a single root node with key equal to 2.
Example 3:

Input: root = [-4,-2,-5]
Output: 0
Explanation: All values are negatives. Return an empty BST.
Example 4:

Input: root = [2,1,3]
Output: 6
Example 5:

Input: root = [5,4,8,3,null,6,3]
Output: 7
```

## Constraints:

* Each tree has at most 40000 nodes..
* Each node's value is between \[-4 *10^4 , 4* 10^4].

## Solutions

1. **postorder traversal with recursion**
2. A tree is not a bst as long as there is one subtree is not bst.

```cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    long res = 0;
    tuple<int, int, long> dfs(TreeNode * root) {
        if (!root) return {INT_MAX, INT_MIN, 0};

        auto [minlv, maxlv, suml] = dfs(root->left);
        auto [minrv, maxrv, sumr] = dfs(root->right);
        if (suml == INT_MIN || sumr == INT_MIN
         || maxlv >= root->val || minrv <= root->val)
            return {0, 0, INT_MIN};

        long cursum = suml + sumr + root->val;
        res = max(res, cursum);
        return {root->left ? minlv : root->val, 
                root->right ? maxrv : root->val,
                cursum};
    }

    int maxSumBST(TreeNode* root) {
        dfs(root);
        return res;
    }
};
`
```