<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 NodeValue{
    public:
    int sum;
    int maxNode;
    int minNode;

    NodeValue(int sum, int maxNode, int minNode)
    {
        this->sum=sum;
        this->maxNode=maxNode;
        this->minNode=minNode;
    }
};

class Solution {
public:
    NodeValue f(TreeNode* root, int &ans)
    {
        if(!root->left && !root->right)
        {
            ans=max(ans,root->val);
            return NodeValue(root->val,root->val,root->val);
        }

        if(!root->left)
        {
            auto right=f(root->right,ans);
            if(root->val<right.minNode)
            {
                ans=max(ans, root->val+right.sum);
                return NodeValue(root->val+right.sum, right.maxNode, root->val);
            }

            return NodeValue(right.sum, INT_MAX, INT_MIN);
        }

        if(!root->right)
        {
            auto left=f(root->left, ans);
            if(root->val>left.maxNode)
            {
                ans=max(ans, root->val+left.sum);
                return NodeValue(root->val+left.sum, root->val, left.minNode);
            }

            return NodeValue(left.sum, INT_MAX, INT_MIN);
        }

        NodeValue left = f(root->left,ans);
        NodeValue right=f(root->right,ans);

        if(left.maxNode<root->val && right.minNode>root->val)
        {
            ans=max(ans, left.sum+root->val+right.sum);
            return NodeValue(left.sum+root->val+right.sum, right.maxNode,left.minNode);
        }

        return NodeValue(max(left.sum,right.sum),INT_MAX,INT_MIN);
    }
    int maxSumBST(TreeNode* root) {
        
        if(!root)
        return 0;
        int ans=0;
        f(root,ans);

        return ans;
        
    }
};
102164