<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 NodeValue {
public:
    int maxNode, minNode, maxSum;
    
    NodeValue(int minNode, int maxNode, int maxSum) {
        this->maxNode = maxNode;
        this->minNode = minNode;
        this->maxSum = maxSum;
    }
};

class Solution {
    int res;
private:
    NodeValue maxSumHelper(TreeNode* root) {
        // An empty tree is a BST of size 0.
        if (!root) {
            return NodeValue(INT_MAX, INT_MIN, 0);
        }
        
        // Get values from left and right subtree of current tree.
        auto left = maxSumHelper(root->left);
        auto right = maxSumHelper(root->right);
        
        // Current node is greater than max in left AND smaller than min in right, it is a BST.
        if (left.maxNode < root->val && root->val < right.minNode) {
            // It is a BST.
            res = max(res,left.maxSum + right.maxSum + root->val);
            return NodeValue(min(root->val, left.minNode), max(root->val, right.maxNode), 
                            left.maxSum + right.maxSum + root->val);
        }
        
        // Otherwise, return [-inf, inf] so that parent can't be valid BST
        return NodeValue(INT_MIN, INT_MAX, max(left.maxSum, right.maxSum));
    }
    public:
    int maxSumBST(TreeNode* root) {
        res = 0;
        int ans = maxSumHelper(root).maxSum;
        return res;
    }
};
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