<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 Info{
    public:
        int minValue;
        int maxValue;
        bool isBST;
        int sum;

        Info(int minValue,int maxValue,bool isBST,int sum){
            this->minValue = minValue;
            this->maxValue = maxValue;
            this->isBST = isBST;
            this->sum = sum;
        }
};

class Solution {
public:
    Info* solve(TreeNode* root,int &maxSum){
        // base case
        if(root == NULL){
            // it is a BST => update maxSum and return INFO
            maxSum = max(maxSum,0);// since currSum = 0;
            return new Info(INT_MAX,INT_MIN,true,0);
        }
        // 1 case me solve remainining recursion will handle
        // L
        Info* leftRecAns = solve(root->left,maxSum);
        // R
        Info* rightRecAns = solve(root->right,maxSum);
        // N
        int minValue = min(min(leftRecAns->minValue,rightRecAns->minValue),root->val);
        int maxValue = max(max(leftRecAns->maxValue,rightRecAns->maxValue),root->val);
        int sum = root->val + leftRecAns->sum + rightRecAns->sum;
        bool isBST = false;
        if(leftRecAns->maxValue < root->val && rightRecAns->minValue > root->val && leftRecAns->isBST && rightRecAns->isBST){
            isBST = true;
            maxSum = max(maxSum,sum);
        }
        return new Info(minValue,maxValue,isBST,sum);
    }

    int maxSumBST(TreeNode* root) {
        int maxSum = INT_MIN;
        solve(root,maxSum);
        return maxSum;
    }
};

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