<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 bound(TreeNode* root, int lower, int upper)
    // {
    //     if(!root) return 1;
    //     return (root->val > lower && root->val < upper) ;
    // }
    int maximum(TreeNode* root,unordered_map<TreeNode* ,int> &mn,unordered_map<TreeNode* ,int> &mx)
    {
        if(!root)
        return INT_MIN;
        if(mx.find(root) != mx.end()) return mx[root];
        return mx[root] = max(root->val,maximum(root->right,mn,mx));
    }
    int minimum(TreeNode* root,unordered_map<TreeNode* ,int> &mn, unordered_map<TreeNode* ,int> &mx)
    {
        if(!root)
        return INT_MAX;
        if(mn.find(root) != mn.end()) return mn[root];
        return mn[root] = min(root->val, minimum(root->left,mn,mx));
    }
    int solve(TreeNode* root, int &maxi,unordered_map<TreeNode* ,int> &mn, unordered_map<TreeNode* ,int> &mx)
    {
        if(!root) return 0;
        int l=solve(root->left,maxi,mn,mx);
        int r=solve(root->right,maxi,mn,mx);
        if(l != INT_MIN && r != INT_MIN && root->val > maximum(root->left,mn,mx) && root->val < minimum(root->right,mn,mx))
        {
            int temp = root->val;
            temp+=l;
            temp+=r;
            maxi=max(maxi,temp);
            return temp;
        }
        else
        return INT_MIN;
    }
    int maxSumBST(TreeNode* root) {
        if(!root) return 0;
        int maxi=0;
        unordered_map<TreeNode*,int> mn;
        unordered_map<TreeNode*, int> mx;
        solve(root,maxi,mn,mx);
        return maxi;
    }
};
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