<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 SubTreeProperties {
public:
    int subteeMinVal;
    int subtreeMaxVal;
    int subtreeSum;
    bool isSubtreeBST;
    
    SubTreeProperties(int subteeMinVal, int subtreeMaxVal, int subtreeSum, bool isSubtreeBST) 
    : subteeMinVal(subteeMinVal), subtreeMaxVal(subtreeMaxVal), subtreeSum(subtreeSum), isSubtreeBST(isSubtreeBST) {}
};

class Solution {
public:
    int maxSumBST(TreeNode* root) {
        int ans = 0;

        // ret: {subteeMinVal, subtreeMaxVal, subtreeSum, isSubtreeBST}
        function<SubTreeProperties*(TreeNode*)> dfs = [&](TreeNode* root) {
            bool canUpdateAns = true;
            int rootVal = root->val;
            int subteeMinVal = rootVal;
            int subtreeMaxVal = rootVal;
            int subtreeSum = rootVal;
            // leaf node can count as single BST
            if (root->left == nullptr && root->right == nullptr) {
                ans = max(ans, subtreeSum);
                return new SubTreeProperties(rootVal, rootVal, subtreeSum, true);
            }
            
            if (root->left != nullptr) {
                SubTreeProperties* leftSubTree = dfs(root->left);
                canUpdateAns &= leftSubTree->subtreeMaxVal < rootVal;
                canUpdateAns &= leftSubTree->isSubtreeBST;
                subtreeSum += leftSubTree->subtreeSum;
                subteeMinVal = min(subteeMinVal, leftSubTree->subteeMinVal);
                subtreeMaxVal = max(subtreeMaxVal, leftSubTree->subtreeMaxVal);
            }

            if (root->right != nullptr) {
                SubTreeProperties* rightSubTree = dfs(root->right);
                canUpdateAns &= rightSubTree->subteeMinVal > rootVal;
                canUpdateAns &= rightSubTree->isSubtreeBST;
                subtreeSum += rightSubTree->subtreeSum;
                subteeMinVal = min(subteeMinVal, rightSubTree->subteeMinVal);
                subtreeMaxVal = max(subtreeMaxVal, rightSubTree->subtreeMaxVal);
            }

            if (canUpdateAns) {
                ans = max(ans, subtreeSum);
            }

            return new SubTreeProperties(subteeMinVal, subtreeMaxVal, subtreeSum, canUpdateAns);
        };
        
        dfs(root);
        return ans;
    }
};

/*
TC: O(n)
SC: O(h)
n = number of nodes in tree, h = tree height

* leaf node count as BST
* if all leaf node < 0 -> return 0 as empty BST


*/
141839