<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 NodeData{
  
    public:
    bool isBst;
    int mn;
    int mx;
    int sum;
    NodeData(){
        mn=INT_MAX;
        mx=INT_MIN;
        isBst=false;
        sum=0;
    }
    
     NodeData(bool bst,int max_ele,int min_ele,int cur_sum){
        mn=min_ele;
        mx=max_ele;
        isBst=bst;
        sum=cur_sum;
    }
    
};

class Solution {
public:
    
    map<TreeNode*,NodeData*>m1;
    void solve(TreeNode* root,int &ans){
       
         
        if(root==NULL){
            return ;
        }
        
        if(root->left==NULL && root->right==NULL){
            
            m1[root]=new NodeData(true,root->val,root->val,root->val);
            ans=max(ans,root->val);
            return ; 
        }
        
        
        solve(root->left,ans);
        solve(root->right,ans);
        
        m1[root]=new NodeData(false,root->val,root->val,root->val);
        
        bool bst=true;
        
        // int cur_sum=root->val;
        if(root->left){
            
            bst=bst&m1[root->left]->isBst;
            
            int left_max=m1[root->left]->mx;
    
            int left_min=m1[root->left]->mn;
            
            if(left_max<root->val){
                m1[root]->mn=left_min;
            }
            else{
                bst=false;
            }
        
            
        }
        
        if(root->right){
            bst=bst&m1[root->right]->isBst;
            
            
            int right_min=m1[root->right]->mn;
            
            int right_max=m1[root->right]->mx;
            
            
            if(right_min>root->val){
                
                m1[root]->mx=right_max;
                
            }
            else{
                bst=false;
            }
            
            
        }
        
        
        if(bst){
        
            int cur_sum=root->val;
            if(root->left){
                cur_sum+=m1[root->left]->sum;
                
            }
            if(root->right){
                cur_sum+=m1[root->right]->sum;
            }
            ans=max(ans,cur_sum);
            m1[root]->sum=cur_sum;
        }        
        m1[root]->isBst=bst;
    
    return;
    }
    
    int maxSumBST(TreeNode* root) {
       int ans=0;
        solve(root,ans);
        return ans;
    }
};
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