**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
```

## View More Similar Problems

## Counting On a Tree

Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n

View Solution →## Polynomial Division

Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie

View Solution →## Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →## The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

View Solution →## Self-Driving Bus

Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever

View Solution →## Unique Colors

You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti

View Solution →