<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) {}
 * };
 */

struct Data {
    int mx;
    int mi;
    int sum;
    bool isBST;
};

class Solution {

    void traverse(TreeNode* root, vector<int>& nums) {
        if (!root)
            return;
        traverse(root->left, nums);
        nums.push_back(root->val);
        traverse(root->right, nums);
    }

    int isValidBST(TreeNode* root) {
        if (!root)
            return 0;
        vector<int> vect;
        traverse(root, vect);
        int sum = vect[0];
        for (int i = 1; i < vect.size(); i++) {
            if (vect[i] <= vect[i - 1])
                return -1;
            sum += vect[i];
        }
        return sum;
    }

    void helper(TreeNode* root, int& ans) {
        if (!root)
            return;
        int k = isValidBST(root);
        // cout<<k<< " root val: "<<root->val<< endl;
        if (isValidBST(root) > 0) {
            ans = max(ans, k);
        }

        helper(root->left, ans);
        helper(root->right, ans);
    }

    void fillData(TreeNode* root, map<TreeNode*, Data>& mp, int &ans) {
        if (!root)
            return;
        if (!root->left && !root->right) {
            Data d = {
                root->val,
                root->val,
                root->val,
                true
            };
            ans = max(ans, root->val);
            mp[root] = d;
            return;
        }
        fillData(root->left, mp, ans);
        fillData(root->right, mp, ans);
        if (!root->left) {
            Data ol = mp[root->right];
            Data ne = {
                max(root->val, ol.mx),
                min(root->val, ol.mi),
                root->val + ol.sum,
                ol.isBST && (ol.mi > root->val),
            };
            mp[root] = ne;
            if (ne.isBST) {
                ans = max(ans, ne.sum);
            }
        } else if (!root->right) {
            Data ol = mp[root->left];
            Data ne = {
                max(root->val, ol.mx),
                min(root->val, ol.mi),
                root->val + ol.sum,
                ol.isBST && (ol.mx < root->val),
            };
            mp[root] = ne;
            if (ne.isBST) {
                ans = max(ans, ne.sum);
            }
        } else {
            Data ol1 = mp[root->left];
            Data ol2 = mp[root->right];
            Data ne = {
                max(root->val, max(ol1.mx, ol2.mx)),
                min(root->val, min(ol1.mi, ol2.mi)),
                root->val + ol1.sum + ol2.sum,
                ol1.isBST && ol2.isBST && (ol1.mx < root->val) &&
                    (ol2.mi > root->val),
            };
            mp[root] = ne;
            if (ne.isBST) {
                ans = max(ans, ne.sum);
            }
        }
        cout << ans << "  "<<mp[root].sum<< " " << mp[root].isBST<< endl;
    }

public:
    int maxSumBST(TreeNode* root) {
        map<TreeNode*, Data> mp;
        int ans = 0;
        fillData(root, mp, ans);
        
        // helper(root, ans);
        // if (ans < 0)
        //     return 0;
        return ans;
    }
};
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