<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 {
public:
    void helper(map<TreeNode*,vector<int>>&mp, TreeNode* root, int& maxVal) {
        if(root == NULL) {
            return;
        }

        if(root->left == NULL && root->right == NULL) {
            mp[root] = {root->val, true, root->val, root->val};
            maxVal = max(maxVal, root->val); // update maxVal for a single node tree
            return;
        }

        if(root->left) {
            helper(mp, root->left, maxVal);
        }
        if(root->right) {
            helper(mp, root->right, maxVal);
        }

        bool isBST = true;
        int sum = root->val, minVal = root->val, maxValInSubtree = root->val;

        if(root->left) {
            if(mp[root->left][1] && mp[root->left][3] < root->val) {
                sum += mp[root->left][0];
                minVal = mp[root->left][2];
            } else {
                isBST = false;
            }
        }

        if(root->right) {
            if(mp[root->right][1] && mp[root->right][2] > root->val) {
                sum += mp[root->right][0];
                maxValInSubtree = mp[root->right][3];
            } else {
                isBST = false;
            }
        }

        mp[root] = {sum, isBST, minVal, maxValInSubtree};
        if(isBST) {
            maxVal = max(maxVal, sum);
        }
    }

    int maxSumBST(TreeNode* root) {
        int maxVal = 0;  // initialize to 0 as we want the maximum sum
        map<TreeNode*,vector<int>> mp;
        helper(mp, root, maxVal);
        return maxVal;
    }
};

201351