<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:
    int maxx = 0;
    unordered_map<TreeNode *, pair<int, int>>hashmap;
    
    int rec(TreeNode *root) {
        int flag = 0, isBST = 0;
        if(root->left && root->right) {
            int left = root->left->val;
            int right = root->right->val;

            flag = min(rec(root->left), rec(root->right));
            
            int a = max(left, max(hashmap[root->left].first, hashmap[root->left].second));
            int b = min(right, min(hashmap[root->right].first, hashmap[root->right].second));

            cout << a << " < " << root->val << " < " << b << endl;
            
            if(a >= root->val || root->val >= b) {
                cout << root->val << endl;
                hashmap[root] = {INT_MAX, INT_MIN};
                isBST = 0;
            }
            else {
                a = min(min(hashmap[root->left].first, hashmap[root->left].second), min(hashmap[root->right].first, hashmap[root->right].second));
                b = max(max(hashmap[root->left].first, hashmap[root->left].second), max(hashmap[root->right].first, hashmap[root->right].second));
                hashmap[root] = {a, b};
                isBST = 1;
            }
            
            if(flag == 1 && isBST == 1) {
                maxx = max(maxx, root->left->val + root->val + root->right->val);
            }
            else {
                isBST = 0;
            }
            root->val += root->left->val + root->right->val;
        }
        else if(root->left) {
            int left = root->left->val;
            
            flag = rec(root->left);

            int a = max(left, max(hashmap[root->left].first, hashmap[root->left].second));

            if(a >= root->val) {
                hashmap[root] = {INT_MAX, INT_MIN};
                isBST = 0;
            }
            else {
                a = min(hashmap[root->left].first, hashmap[root->left].second);
                hashmap[root] = {a, root->val};
                isBST = 1;
            }
            
            if(flag == 1 && isBST == 1) {
                maxx = max(maxx, root->left->val + root->val);
            }
            else {
                isBST = 0;
            }
            root->val += root->left->val;
        }
        else if(root->right) {
            int right = root->right->val;
            
            flag = rec(root->right);

            int b = min(right, min(hashmap[root->right].first, hashmap[root->right].second));
            
            if(root->val >= b) {
                hashmap[root] = {INT_MAX, INT_MIN};
                isBST = 0;
            }
            else {
                b = max(hashmap[root->right].first, hashmap[root->right].second);
                hashmap[root] = {root->val, b};
                isBST = 1;
            }
            
            if(flag == 1 && isBST == 1) {
                maxx = max(maxx, root->right->val + root->val);
            }
            else {
                isBST = 0;
            }
            root->val += root->right->val;
        }
        else {
            isBST = 1;
            hashmap[root] = {root->val, root->val};
            maxx = max(maxx, root->val);
        }

        return isBST;
    }

    int maxSumBST(TreeNode* root) {
        rec(root);

        return maxx;
    }
};
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