<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:
    class Box {
        public:
        bool isBST;
        int sum;
        int min;
        int max;

        Box () {
            isBST = false;
            int sum = 0;
            int min = -1;
            int max = -1;
        }
    };

    Box* solve (TreeNode* root, int& maxSum) {

        // Leaf Node
        if (!root->left && !root->right) {
            Box* newBox = new Box ();
            newBox->isBST = true;
            newBox->sum = root->val;
            newBox->min = root->val;
            newBox->max = root->val;
            maxSum = max (newBox->sum, maxSum);
            return newBox;
        }
        
        // Right Subtree does not Exist.
        else if (root->left && !root->right) {
            Box* leftBox = solve (root->left, maxSum);
            Box* newBox = new Box ();

            if (leftBox->isBST && root->val > leftBox->max) {
                newBox->isBST = true;
                newBox->sum = leftBox->sum + root->val;
                newBox->min = leftBox->min;
                newBox->max = root->val;
                maxSum = max (newBox->sum, maxSum);
            }
            return newBox;
        }

        // Left Subtree does not Exist.
        else if (root->right && !root->left) {
            Box* rightBox = solve (root->right, maxSum);
            Box* newBox = new Box ();

            if (rightBox->isBST && root->val < rightBox->min) {
                newBox->isBST = true;
                newBox->sum = rightBox->sum + root->val;
                newBox->min = root->val;
                newBox->max = rightBox->max;
                maxSum = max (newBox->sum, maxSum);
            }
            return newBox;
        }

        // Both Subtrees Exist
        else {
            Box* leftBox = solve (root->left, maxSum);
            Box* rightBox = solve (root->right, maxSum);
            Box* newBox = new Box ();

            if (leftBox->isBST && rightBox->isBST && root->val > leftBox->max && root->val < rightBox->min) {
                newBox->isBST = true;
                newBox->sum = leftBox->sum + rightBox->sum + root->val;
                newBox->min = leftBox->min;
                newBox->max = rightBox->max;
                maxSum = max (newBox->sum, maxSum);
            }
            return newBox;  
        }

        return nullptr;
    }
    int maxSumBST(TreeNode* root) {
        int maxSum = 0;

        solve (root, maxSum);

        return maxSum;
    }
};
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