<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:
    int maxi=INT_MIN;
    
    vector<int> f(TreeNode *root){
        if(root){
            if(root->left==NULL && root->right==NULL){
                maxi=max(maxi,root->val);
                return {1,root->val,root->val,root->val};
            }
            auto v1 = f(root->left);
            auto v2 = f(root->right);
            int curr_val = root->val;
            bool isLeftBST = v1[0];
            bool isRightBST = v2[0];
            bool isCurrBST = false;
            int curr_sum=0;
            if(isLeftBST && isRightBST){
                int max_left = v1[1];
                int min_right = v2[2];
                if(curr_val>max_left && curr_val<min_right) isCurrBST = true;
            }
            if(isCurrBST){
                curr_sum = v1[3]+v2[3]+curr_val;
                maxi=max(maxi,curr_sum);
                cout<<maxi<<endl;
                return {1,v2[1]==INT_MIN?curr_val:v2[1],v1[2]==INT_MAX?curr_val:v1[2],curr_sum};
            }
            return {0,0,0,0};
        }
        return {1,INT_MIN,INT_MAX,0};
    }

    int maxSumBST(TreeNode* root) {
        f(root);
        return max(maxi,0);
    }   
};
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