<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  solve(TreeNode* root,int &count,map<int,pair<int,int>>&mp)
  {
    int a=0;
    int b=0;
    int d=root->val;
    int dxx,dyy;
    if(root->left!=NULL)
    {
         a=solve(root->left,count,mp);
       
    }
    if(root->right!=NULL)
    {
        b=solve(root->right,count,mp);
    }

    if(root->left==NULL&&root->right==NULL)
    {
            d=root->val;
           mp[root->val]={root->val,root->val};
    }
     else if(root->left==NULL&&root->right!=NULL)
     {
         if(mp[root->right->val].first>root->val)
         {
              if(b!=-1e9)
              {
                dxx=min(root->val,mp[root->right->val].first);
                dyy=max(root->val,mp[root->right->val].second);
                mp[root->val]={dxx,dyy};
                 d=d+b;
              }
              else{
                d=-1e9;
              }
         }
         else{
            d=-1e9;
         }
     }
     else if(root->left!=NULL&&root->right==NULL)
     {
         if(mp[root->left->val].second<root->val)
         {
              if(a!=-1e9)
              {
                dxx=min(root->val,mp[root->left->val].first);
                dyy=max(root->val,mp[root->left->val].second);
                mp[root->val]={dxx,dyy};
                 d=d+a;
              }
              else{
                d=-1e9;
              }
         }
         else{
            d=-1e9;
         }
     }
else if(root->left!=NULL&&root->right!=NULL)
     {
         
         if(mp[root->left->val].second<root->val&&mp[root->right->val].first>root->val)
         {
              if(a!=-1e9&&b!=-1e9)
              {

                dxx=min(root->val,mp[root->left->val].first);
                dyy=max(root->val,mp[root->left->val].second);
                dxx=min(dxx,mp[root->right->val].first);
                dyy=max(dyy,mp[root->right->val].second);

                 mp[root->val]={dxx,dyy};
                 d=d+a+b;
              }
              else{
                d=-1e9;
              }
         }
         else{
            d=-1e9;
         }
     }

   
    count=max(count,d);

    return d;
    
  }
    int maxSumBST(TreeNode* root) {

        int count=0;
    
       map<int,pair<int,int>>mp;
    int d= solve(root,count,mp);

       return count;
    }
};
125969