<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 Box {
public:
    bool bst;
    int size;
    int min, max;
    int sum;  // To store the sum of nodes in the current subtree

    Box(int data) {
        bst = true;
        size = 1;
        min = data;
        max = data;
        sum = data;
    }
};

class Solution {
public:
    Box* find(TreeNode *root, int &maxSum) {
        // Base case: when the node is NULL
        if (root == NULL) {
            return NULL;
        }

        // Leaf node
        if (root->left == NULL && root->right == NULL) {
            maxSum = max(maxSum, root->val);
            return new Box(root->val);
        }

        // Left child only
        if (root->left != NULL && root->right == NULL) {
            Box* leftBox = find(root->left, maxSum);
            if (leftBox->bst && leftBox->max < root->val) {
                leftBox->size++;
                leftBox->max = root->val;
                leftBox->sum += root->val;
                maxSum = max(maxSum, leftBox->sum);
                return leftBox;
            } else {
                leftBox->bst = false;
                return leftBox;
            }
        }

        // Right child only
        if (root->left == NULL && root->right != NULL) {
            Box* rightBox = find(root->right, maxSum);
            if (rightBox->bst && rightBox->min > root->val) {
                rightBox->size++;
                rightBox->min = root->val;
                rightBox->sum += root->val;
                maxSum = max(maxSum, rightBox->sum);
                return rightBox;
            } else {
                rightBox->bst = false;
                return rightBox;
            }
        }

        // Both children exist
        Box* leftBox = find(root->left, maxSum);
        Box* rightBox = find(root->right, maxSum);
        if (leftBox->bst && rightBox->bst && leftBox->max < root->val && rightBox->min > root->val) {
            Box* currentBox = new Box(root->val);
            currentBox->size = leftBox->size + rightBox->size + 1;
            currentBox->min = leftBox->min;
            currentBox->max = rightBox->max;
            currentBox->sum = leftBox->sum + rightBox->sum + root->val;
            maxSum = max(maxSum, currentBox->sum);
            return currentBox;
        } else {
            Box* invalidBox = new Box(root->val);
            invalidBox->bst = false;
            return invalidBox;
        }
    }

    /*You are required to complete this method */
    // Return the sum of the largest sub-tree which is also a BST
    int maxSumBST(TreeNode *root) {
        int maxSum = 0;
        find(root, maxSum);
        return maxSum;
    }
};

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