<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 {
    unordered_map<TreeNode*, int> rtc;
    int rightt(TreeNode* r) {
        if (!r->right)
            return rtc[r] = r->val;

        if (rtc.count(r))
            return rtc[r];

        return rtc[r] = rightt(r->right);
    }

    unordered_map<TreeNode*, int> ltc;
    int leftt(TreeNode* r) {
        if (!r->left)
            return ltc[r] = r->val;
        if (ltc.count(r))
            return ltc[r];

        return ltc[r] = leftt(r->left);
    }

public:
    int maxSumBST(TreeNode* root) {
        int res = 0;

        function<pair<bool, int>(TreeNode*)> dfs = [&](TreeNode* p) {
            if (!p)
                return (pair<bool, int>){true, 0};

            auto [okl, left] = dfs(p->left);
            auto [okr, right] = dfs(p->right);

            int sum = left + right + p->val;

            if (p->left && (!okl || rightt(p->left) >= p->val))
                return (pair<bool, int>){false, sum};
            if (p->right && (!okr || leftt(p->right) <= p->val))
                return (pair<bool, int>){false, sum};

            res = max(sum, res);
            return (pair<bool, int>){true, sum};
        };

        dfs(root);

        return res;
    }
};
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