<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> checkBST(TreeNode* root,int &ans){
       if(!root){
        return {INT_MIN,INT_MAX,0};
       }

       vector<int> lft = checkBST(root->left,ans);
       vector<int> rht = checkBST(root->right,ans);

       if(!root->left&&!root->right){
        vector<int> temp(3,0);
        temp[0] = temp[1] = temp[2] = root->val;
        ans = max(ans,root->val);
        return temp;
       }
       
       int sum = 0;
       if(!root->left){
         if(rht[1] > root->val && rht[0] > root->val){
            sum = rht[2] + root->val;
            ans = max(ans,sum);
            rht[0] = root->val;
            rht[2] = sum;
            return rht;
         }
         else{
            vector<int> temp = {INT_MAX,INT_MIN,0};
            return temp;
         }
       }

       if(!root->right){
        if(lft[0] < root->val && lft[1] < root->val){
           sum = lft[2] + root->val;
           ans = max(ans,sum);
           lft[1] = root->val;
           lft[2] = sum;
           return lft;
        }
        else{
            vector<int> temp = {INT_MAX,INT_MIN,0};
            return temp;
        }
       }

       if(lft[0] < root->val && lft[1] < root->val && rht[1] > root->val && rht[0] > root->val){
        sum = lft[2] + rht[2] + root->val;
        ans = max(ans,sum);
        lft[1] = rht[1];
        lft[2] = sum;
        return lft;
       }
       else{
         return {INT_MAX,INT_MIN,0};
       }
    }

    int maxSumBST(TreeNode* root) {
        int ans = 0;
        vector<int> temp = checkBST(root,ans);
        return ans;
    }
};
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