<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 root_data{
    public:
    int a, b, c, d, e;
    root_data(){
        a = 0; 
        b = INT_MIN;
        c = INT_MAX;
        d = INT_MIN;
        e = INT_MIN;
    }
};
class Solution {
public:
    
    root_data helper (TreeNode*root) {
        root_data ans;
        if (root==NULL) {
            ans.a = -1;
            ans.b = 0;
            ans.c = 0;
            ans.d = 0;
            ans.e = 0;
            return ans;
        }
        root_data left_data = helper(root->left);
        root_data right_data = helper(root->right);
        
        if (left_data.a==0 || right_data.a==0) {
            ans.a = 0;
            ans.d = max({left_data.d, right_data.d});
        }
        else if ((left_data.a!=-1 && left_data.c>=root->val) || (right_data.a!=-1 && right_data.b<=root->val)) {
            ans.a = 0;
            ans.b = 0; 
            ans.c = 0; 
            ans.d = max({left_data.d, right_data.d});
            ans.e = 0;
        }
        else if (left_data.a==-1 && right_data.a==-1){
            ans.a = 1;
            ans.b = root->val;
            ans.c = root->val;
            ans.d = root->val;
            ans.e = root->val;
        }
        else if (left_data.a==1 && right_data.a==1){
            ans.a = 1;
            ans.b = left_data.b;
            ans.c = right_data.c;
            ans.e = root->val + left_data.e + right_data.e;
            ans.d = max({ans.e, left_data.d, right_data.d});
        }
        else if (left_data.a==1) {
            ans.a = 1;
            ans.b = left_data.b;
            ans.c = root->val;
            ans.e = root->val + left_data.e;
            ans.d = max({ans.e, left_data.d});
        }
        else{
            ans.a = 1;
            ans.b = root->val;
            ans.c = right_data.c;
            ans.e = root->val + right_data.e;
            ans.d = max({ans.e, right_data.d});
        }

        cout<<"root->val: "<<root->val<<endl;
        cout<<ans.a<<" "<<ans.b<<" "<<ans.c<<" "<<ans.d<<" "<<ans.e<<endl;

        return ans;

    }

    int maxSumBST(TreeNode* root) {
        // we will try to handle this problem using return values
        // [a, b, c, d, e]
        // a--> that whether or not the subtree is bst 
        // b--> min value in the btree
        // c--> max value in the btree
        // d--> max_ans satisfying the condtions found inside the tree
        // e--> max_ans satisfying the condtions including the current root
        root_data ans = helper(root);
        return max(ans.d, 0);
    }
};
122001