<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>

class Solution {
    int maxSum = INT_MIN;

    vector<int> findMaxSumBST(TreeNode* rootNode) {
        if (!rootNode)
            return {true, 0, INT_MAX, INT_MIN}; // {isBST, sum, minVal, maxVal}

        auto BSTInfoLeft  = findMaxSumBST(rootNode->left);
        auto BSTInfoRight = findMaxSumBST(rootNode->right); 

        bool isLeftBST  = (BSTInfoLeft[0] && BSTInfoLeft[3] < rootNode->val);
        bool isRightBST = (BSTInfoRight[0] && BSTInfoRight[2] > rootNode->val);

        int BSTSum = BSTInfoLeft[1] + BSTInfoRight[1] + rootNode->val;

        if (isLeftBST && isRightBST)
            maxSum = max(maxSum, BSTSum);

        int minVal = min(rootNode->val, BSTInfoLeft[2]);
        int maxVal = max(rootNode->val, BSTInfoRight[3]);

        return {isLeftBST && isRightBST, BSTSum, minVal, maxVal};
    }

public:
    int maxSumBST(TreeNode* rootNode) {
        findMaxSumBST(rootNode);
        return (maxSum <= 0) ? 0 : maxSum;
    }
};
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