<p>Given a <strong>binary tree</strong> <code>root</code>, return <em>the maximum sum of all keys of <strong>any</strong> sub-tree which is also a Binary Search Tree (BST)</em>.</p>

<p>Assume a BST is defined as follows:</p>

<ul>
	<li>The left subtree of a node contains only nodes with keys <strong>less than</strong> the node&#39;s key.</li>
	<li>The right subtree of a node contains only nodes with keys <strong>greater than</strong> the node&#39;s key.</li>
	<li>Both the left and right subtrees must also be binary search trees.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/01/30/sample_1_1709.png" style="width: 320px; height: 250px;" /></p>

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

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/01/30/sample_2_1709.png" style="width: 134px; height: 180px;" /></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [-4,-2,-5]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All values are negatives. Return an empty BST.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 4 * 10<sup>4</sup>]</code>.</li>
	<li><code>-4 * 10<sup>4</sup> &lt;= Node.val &lt;= 4 * 10<sup>4</sup></code></li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxSumBST(TreeNode* root) {
        int ans=0;
        vector<int>val= maxsumbst(root, ans);
        return ans;
    }
    vector<int> maxsumbst(TreeNode* root, int &ans){
        if(root==NULL) return {};           // sum         minval,  maxval
        if(!root->left&&!root->right) return {root->val, root->val, root->val, true};
        vector<int> l=maxsumbst(root->left, ans);
        vector<int> r=maxsumbst(root->right, ans);
        if(l.size()==0){
            ans=max(ans, r[0]);
            if(!r[3]||r[1]<=root->val) return {r[0], INT_MAX, INT_MIN, false};
            ans=max(ans, root->val+r[0]);
            return {root->val+r[0], root->val, r[2], true};
        }
        else if(r.size()==0){
            ans=max(ans, l[0]);
            if(!l[3]||l[2]>=root->val) return {l[0], INT_MAX, INT_MIN, false};
            ans=max(ans, root->val+l[0]);
            return {root->val+l[0], l[1], root->val, true};
        }
        ans=max(ans, max(l[0], r[0]));
        if(!l[3]||!r[3]||(l[2]>=root->val||r[1]<=root->val)) return {max(l[0], r[0]), INT_MAX, INT_MIN, false};
        // isvalid BST
        ans=max(ans, l[0]+root->val+r[0]);
        return {l[0]+root->val+r[0], l[1], r[2], true}; 
    }
};
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