<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>

#include <climits> // For INT_MIN and INT_MAX

class Info {
public:
    int maxi;
    int mini;
    int sum;
    bool isBst;

    Info(int mx, int mn, int s, bool x) {
        maxi = mx;
        mini = mn;
        sum = s;
        isBst = x;
    }
};

class Solution {
public:
    int ans = 0;

    Info f(TreeNode *root) {
        if (!root)
            return Info(INT_MIN, INT_MAX, 0, true); // Adjusted for null node

        Info left = f(root->left);
        Info right = f(root->right);

        // Check if current tree is a BST
        if (left.isBst && right.isBst && root->val > left.maxi && root->val < right.mini) {
            int currentSum = left.sum + right.sum + root->val;
            ans = max(ans, currentSum);
            return Info(max(root->val, right.maxi), min(root->val, left.mini), currentSum, true);
        }

        // If it's not a BST
        return Info(0, 0, 0, false);
    }

    int maxSumBST(TreeNode* root) {
        f(root);
        return ans;
    }
};
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