# Largest Binary Search Subtree in Nodes - Amazon Top Interview Questions

### Problem Statement :

```Given a binary tree root, find the largest subtree (the one with the most nodes) that is a binary search tree.

Constraints

n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [1, [3, [2, null, null], [5, null, null]], null]

Output

[3, [2, null, null], [5, null, null]]

Explanation

The root is not a valid binary search tree, but the tree beginning at 3 is.```

### Solution :

```                        ```Solution in C++ :

tuple<int, int, int, bool> solve(Tree *root, int &mx, Tree *&res) {
// for null node
if (!root) return {INT_MAX, INT_MIN, 0, true};

// fetch answer for the subtrees starting at the left and the right subtree
auto [leftMin, leftMax, leftRes, leftValid] = solve(root->left, mx, res);
auto [rightMin, rightMax, rightRes, rightValid] = solve(root->right, mx, res);

/*
* check the validity of BST for the subtree at the current node.
* resHere = 1 + leftRes + rightRes; iff validHere == true;
* else 0
*/
int val = root->val;
bool validHere = leftValid && rightValid && leftMax < val && rightMin > val;
int resHere = validHere * (1 + leftRes + rightRes);

// if better answer found update
if (mx < resHere) {
mx = resHere;
res = root;
}

return {min({leftMin, rightMin, val}), max({leftMax, rightMax, val}), resHere, validHere};
}

Tree *solve(Tree *root) {
int mx = 0;
Tree *res = nullptr;
solve(root, mx, res);

return res;
}```
```

```                        ```Solution in Java :

import java.util.*;

/**
* public class Tree {
*   int val;
*   Tree left;
*   Tree right;
* }
*/
class Solution {
Tree ret = null;
int biggest_size = 0;
int size = 0;
public Tree solve(Tree root) {
recurse(root);
return ret;
}
public void recurse(Tree root) {
if (root == null)
return;
if (is_binary(root) && size > biggest_size) {
ret = root;
biggest_size = size;
}
size = 0;
recurse(root.left);
recurse(root.right);
}
public boolean is_binary(Tree root) {
if (root == null)
return true;
if (root.left != null && root.left.val > root.val) {
return false;
}
if (root.right != null && root.right.val < root.val) {
return false;
}
size++;
return is_binary(root.left) && is_binary(root.right);
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, root):
numNodes = {}
# getting the number of Nodes of each subtree
def dfs(root):
if not root:
return 0
l, r = dfs(root.left), dfs(root.right)
numNodes[root] = 1 + l + r
return 1 + l + r

dfs(root)

storeMax, retRoot = 0, None
# validating the BST from bottom to top while tracking the maxNum nodes
def validate(root, parent):
nonlocal storeMax, retRoot
if not root:
return [True, (parent, parent)]
resL, minMaxL = validate(root.left, root)
resR, minMaxR = validate(root.right, root)
# parent BST ? ==> both left and right are BST and they satisfy the range property
resRoot = resL and resR and minMaxL[0].val <= root.val <= minMaxR[1].val
if resRoot:
if numNodes[root] > storeMax:
retRoot, storeMax = root, numNodes[root]
return [True, (minMaxL[1], minMaxR[0])]
# current subtree not a BST ==> from this side there will be no contribution towards the result
return [False, ((None), (None))]

validate(root, None)
return retRoot```
```

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