<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 Node {
public:
    int sum;
    int left;
    int right;
    Node(int _s, int _l, int _r){
        sum = _s;
        left = _l;
        right = _r;
    }
};
class Solution {
public:
    Node* inorder (TreeNode* root,int& ans){
        if(!root)   return nullptr;
        
        Node* lnode = inorder(root->left,ans);
        Node* rnode = inorder(root->right,ans);

        int sum = root->val;
        int left = root->val;
        int right = root->val;
        if(lnode){
            sum+= lnode->sum;
            left = lnode->left;
        }
        if(rnode){
            sum+= rnode->sum;
            right = rnode->right;
        }
        if((!lnode || lnode->right < root->val )&&(!rnode || root->val < rnode->left)){
            // cout<<root->val<<":"<<sum<<endl;
            ans = max(ans,sum);
            return new Node(sum,left,right);
        }
        return new Node(-1e6,INT_MIN,INT_MAX);
    }
    int maxSumBST(TreeNode* root) {
        int ans = 0;
        inorder(root,ans);
        return ans;
    }
};
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