<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:
    struct Node {
      public:
        int size;
        int max;
        int min;
        int sum;

        Node(int val) {
            size = -1;
            max = val;
            min = val;
            sum = val;
        }   

        Node(int s, int mn, int mx, int sm) {
            size = s;
            min = mn;
            max = mx;
            sum = sm;
        }
    };

    int largest = 0;

    Node* f(TreeNode* root){
        

        if(root == NULL) {
            return NULL; 
        }

        Node* left = f(root->left);
        Node* right = f(root->right);

        Node* a = new Node(root->val); 

        if(left == NULL && right == NULL) {
            a->size = 1; 
            largest = max(largest , a->sum);
            cout<<largest<< " ";
            return a;
        }
        else if(left == NULL) {
            if(root->val < right->min && right->size != -1) {
                a->size = 1 + right->size;
                a->min = root->val;
                a->max = right->max;
                a->sum += right->sum;
                largest = max(largest,a->sum);
                cout<<largest<< " ";
                return a;
            }
            cout<<largest<< " ";
            return a;
        }
        else if(right == NULL){
            if(root->val > left->max && left->size != -1) {
                a->size = 1 + left->size;
                a->min = left->min;
                a->max = root->val;
                a->sum += left->sum;
                largest = max(largest,a->sum);
                cout<<largest<< " ";
                return a;
            }
            cout<<largest<< " ";
            return a;
        }
        else if(root->val > left->max && root->val < right->min && right->size != -1 && left->size != -1){
            a->size = 1 + left->size + right->size;
            a->min = left->min;
            a->max = right->max;
            a->sum += left->sum + right->sum;
            largest = max(largest, a->sum);
            cout<<largest<< " ";
            return a;
        }
        else {
            cout<<largest<< " ";
            return a;
        }
        
    }

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