<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) {}
 * };
 */
 auto speedUp = []() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 0;
}();

struct validAns{
    int mnValue ;
    int mxValue ;
    int sum = 0;
    bool valid = true;
    validAns(){
        valid = true;
        mnValue = INT_MAX;
        mxValue = INT_MIN;
        sum = 0;
    }
    validAns(bool valid, int sum, int mxVal, int mnVal){
        this->valid = valid;
        this->sum = sum;
        mnValue = mnVal;
        mxValue = mxVal;
    }
};
class Solution {
public:
    validAns* dfs(TreeNode* root, int &ans){
        if(!root) return new validAns();
        validAns *leftTree = dfs(root->left, ans), *rightTree = dfs(root->right, ans);
        bool ansValidity = (leftTree->valid && rightTree->valid);
        int sum = leftTree->sum + rightTree->sum + root->val;
        // cout<<sum<<endl;
        // if(root->right && root->right->val <= root->val ) {
        //     ansValidity = false;
        // }
        int mxVal = max({root->val, leftTree->mxValue, rightTree->mxValue});
        int mnVal = min({root->val, leftTree->mnValue, rightTree->mnValue});
        if(leftTree->mxValue >= root->val) ansValidity = false;
        // if(root->left && root->left->val >= root->val){
        //     ansValidity = false;
        // }
        if(rightTree->mnValue <= root->val) ansValidity = false;
        if(ansValidity){
            ans = max(ans, sum);
        }
        return new validAns(ansValidity, sum, mxVal , mnVal );
    }
    int maxSumBST(TreeNode* root) {
        int ans = 0;
        dfs(root, ans);
        return ans;
    }
};
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