<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 Node{
    public:
    int sum;
    int minval;
    int maxval;
    Node* left;
    Node* right;
    bool isBST;

    Node(){
        sum=0;
        minval = INT_MAX;
        maxval = INT_MIN;
        left = NULL;
        right = NULL;
        isBST = true;
    }

};
class Solution {
public:
 int ans;

    Node* create(TreeNode* root){
        if(root==NULL) return NULL;
        Node* temp = new Node();
        temp->sum = root->val;
        bool isbst = true;
        if(root->left){
            temp->left = create(root->left);
            if(temp->left->isBST==false || temp->left->maxval >= root->val) isbst = false;
            temp->sum+=temp->left->sum;
            temp->minval = min(temp->minval,temp->left->minval);
            temp->maxval = max(temp->maxval,temp->left->maxval);
        }

        if(root->right){
            temp->right = create(root->right);
            if(!temp->right->isBST || temp->right->minval <= root->val) isbst = false;
            temp->sum+=temp->right->sum;
            temp->minval = min(temp->minval, temp->right->minval);
            temp->maxval = max(temp->maxval, temp->right->maxval);
        }

        temp->minval = min(temp->minval, root->val);
        temp->maxval = max(temp->maxval,root->val);
        temp->isBST = isbst;
        return temp;
    }

    void addsum(Node* root){
        if(root==NULL) return;
        if(root->isBST) ans=max(ans,root->sum);
        addsum(root->left);
        addsum(root->right);
    }
    int maxSumBST(TreeNode* root) {
        Node* rot = create(root);
        ans = 0;
        addsum(rot);
        return ans;
    }
};
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