<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>

class Solution {
public:
    int maxSumBST(TreeNode* root) {
        if(!root) return 0;
        if(root->left == NULL && root->right == NULL) return root->val;
        int maxSum = 0;
        int sum = 0;
        int max = INT_MIN;
        bool isBST;

        if(root->left && root->left->val > root->val || root->right && root->right->val < root->val){
            isBST = false;
        }
        else
            isBST = true;

        int a=0;
        int b=0;
        dfs(root, sum, max, isBST,a,b);
        if(max < 0) return 0;
        return max;
    }
    void dfs(TreeNode* root, int& sum, int& max, bool& isBST,int& Max , int& Min ){
        if(!root){
            isBST = true;
            Max = 0;
            Min = 0;
            return ;
        }

        dfs(root->left, sum, max, isBST, Max, Min);
        int leftSum = sum;
        int leftMax = Max;
        int leftMin = Min;
        bool leftIsBST = isBST;

        sum = 0;
        dfs(root->right, sum, max, isBST, Max, Min);
        int rightSum = sum;
        int rightMax = Max;
        int rightMin = Min;
        bool rightIsBST = isBST;


        if(leftIsBST && rightIsBST){
            if(root->left && root->left->val >= root->val || root->right && root->right->val <= root->val){
                isBST = false;
                return ;
            }
            isBST = true;
            if(leftMax ==0){
                leftMax = INT_MIN;
            }
            if(rightMin == 0){
                rightMin = INT_MAX;
            }
            if(leftMax < root->val && rightMin > root->val){
                sum = leftSum + rightSum + root->val;
                max = max > sum ? max : sum;
            }
            else{
                isBST = false;
            }

        }
        else{
            isBST = false;
        }
        Max = getthemaxminof3(root->val, leftMax, rightMax, true);
        Min = getthemaxminof3(root->val, leftMin, rightMin, false);


    }
    int getthemaxminof3(int a, int b, int c ,bool mode){
        if(mode){
            return a > b ? (a > c ? a : c) : (b > c ? b : c);
        }
        else{
            return a < b ? (a < c ? a : c) : (b < c ? b : c);
        }
    }
};
106131