<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:
    vector<int> isBst(TreeNode *root, int &ans) {
        // return vector structure => {isBST, sum, minval of subtree, maxval of subtree}
        
        if(!root) {
            return vector<int> {1, 0, INT_MAX, INT_MIN}; // valid bst but data value returned to parent should be 0
        }

        if(!root->left && !root->right) {
            ans = max(ans, root->val);
            return vector<int>{1, root->val, root->val, root->val};
        }

        vector<int> isLeftSubtreeBst = isBst(root->left, ans);
        vector<int> isRightSubtreeBst = isBst(root->right, ans);

        vector<int> ret(4);
        if(isLeftSubtreeBst[0] && isRightSubtreeBst[0] && root->val > isLeftSubtreeBst[3] && root->val < isRightSubtreeBst[2]) {
            ret[0] = 1;
            ret[1] = isLeftSubtreeBst[1] + isRightSubtreeBst[1] + root->val;
            ret[2] = min(root->val, isLeftSubtreeBst[2]);
            ret[3] = max(root->val, isRightSubtreeBst[3]);
            ans = max(ans, ret[1]);
        } else {
            ret[0] = 0;
            ret[1] = 0;
            ret[2] = INT_MIN;
            ret[3] = INT_MAX;
        }
        return ret;
    }

    int maxSumBST(TreeNode* root) {
        int ans = INT_MIN;   
        vector<int> res =  isBst(root, ans);   
        if(ans > 0)
            return ans;
        return 0;
    }
};
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