<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:
    vector<int> check(TreeNode* root,int& ans){
        if(root == NULL) return {0,0,INT_MIN};

        if(root->left == NULL && root->right == NULL){ ans = max(root->val,ans); return {root->val,root->val,root->val};}

        vector<int> left = check(root->left,ans);
        vector<int> right = check(root->right,ans);

        if(left[0] != INT_MIN && right[0] != INT_MIN){
            if( ((left[0] < root->val && left[1] < root->val) || left[2] == INT_MIN) && ((right[0] > root->val && right[1] > root->val) || right[2] == INT_MIN) ){
                int leftS = left[2] == INT_MIN ? 0 : left[2];
                int rightS = right[2] == INT_MIN ? 0 : right[2];
                int curMax = leftS+rightS+root->val;
                ans  = max(ans,curMax);
                int mini = root->val;
                int maxi = root->val;
                if(left[2] != INT_MIN || right[2] != INT_MIN){
                    if(left[2] != INT_MIN){
                        mini = min(left[0],mini);
                        maxi = max(left[1],maxi);
                    }
                    if(right[2] != INT_MIN){
                        mini = min(right[0],mini);
                        maxi = max(right[1],maxi);
                    }
                }
                return {mini,maxi,curMax};
            }else return {INT_MIN,INT_MIN,INT_MIN};
        }else return {INT_MIN,INT_MIN,INT_MIN};
    }
    int maxSumBST(TreeNode* root) {
        int ans = 0;
        check(root,ans);
        if(ans < 0) return 0;
        return ans;
    }
};
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