**Largest Binary Search Subtree in Value - Amazon Top Interview Questions**

### Problem Statement :

Given a binary tree root, return the largest sum of a subtree that is also a binary search tree. Constraints 0 ≤ n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [1, [5, [6, null, null], [7, null, null]], [3, [2, null, null], [4, null, null]]] Output 9 Explanation The binary search tree rooted at 3 has the largest sum with 3 + 2 + 4 = 9

### Solution :

` ````
Solution in C++ :
pair<pair<bool, int>, pair<int, int>> helper(Tree* root, int& res) {
if (!root) {
return {{1, 0}, {INT_MAX, INT_MIN}};
}
pair<pair<bool, int>, pair<int, int>> leftans = helper(root->left, res);
pair<pair<bool, int>, pair<int, int>> rightans = helper(root->right, res);
pair<pair<bool, int>, pair<int, int>> ans;
ans.first.first = leftans.first.first && rightans.first.first &&
root->val > leftans.second.second && root->val < rightans.second.first;
ans.first.second = leftans.first.second + rightans.first.second + root->val;
ans.second.first = min({root->val, leftans.second.first, rightans.second.first});
ans.second.second = max({root->val, leftans.second.second, rightans.second.second});
if (ans.first.first) {
res = max(res, ans.first.second);
}
return ans;
}
int solve(Tree* root) {
int res = 0;
helper(root, res);
return res;
}
```

` ````
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
private int maxSubtreeSum = 0;
static class Node {
int sum;
int min;
int max;
boolean isBST;
public Node(int sum, int min, int max, boolean isBST) {
this.sum = sum;
this.min = min;
this.max = max;
this.isBST = isBST;
}
}
public int solve(Tree root) {
dfs(root);
return maxSubtreeSum;
}
private Node dfs(Tree node) {
if (node.left == null && node.right == null) {
maxSubtreeSum = Math.max(maxSubtreeSum, node.val);
return new Node(node.val, node.val, node.val, true);
}
int nodeVal = node.val;
boolean isBST = true;
int sum, minval, maxval;
sum = nodeVal;
minval = nodeVal;
maxval = nodeVal;
if (node.left != null) {
Node lnode = dfs(node.left);
isBST = (isBST && lnode.isBST && (nodeVal > lnode.max));
sum += lnode.sum;
minval = Math.min(minval, lnode.min);
// maxval = Math.max(maxval, lnode.max);
}
if (node.right != null) {
Node rnode = dfs(node.right);
isBST = (isBST && rnode.isBST && nodeVal < rnode.min);
sum += rnode.sum;
maxval = Math.max(maxval, rnode.max);
// minval = Math.min(minval, rnode.min);
}
if (isBST) {
maxSubtreeSum = Math.max(maxSubtreeSum, sum);
return new Node(sum, minval, maxval, true);
}
return new Node(sum, minval, maxval, false);
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, root):
ans = 0
def go(root):
nonlocal ans
if not root:
return (0, 1e9, -1e9, 1)
ls, lmin, lmax, isl = go(root.left)
rs, rmin, rmax, isr = go(root.right)
if not isl or not isr or (root.val <= lmax) or (root.val >= rmin):
return (0, 0, 0, 0)
ans = max(ans, root.val + ls + rs)
return (root.val + ls + rs, min(root.val, lmin), max(root.val, rmax), 1)
go(root)
return ans
```

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