# Largest BST in a Binary Tree | Set 2

Given a Binary Tree, write a function that returns the size of the largest subtree which is also a Binary Search Tree (BST). If the complete Binary Tree is BST, then return the size of whole tree.

Examples:

```Input:
5
/  \
2    4
/  \
1    3

Output: 3
The following subtree is the
maximum size BST subtree
2
/  \
1    3

Input:
50
/    \
30       60
/  \     /  \
5   20   45    70
/  \
65    80
Output: 5
The following subtree is the
maximum size BST subtree
60
/  \
45    70
/  \
65    80
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We have discussed a two methods in below post.
Find the largest BST subtree in a given Binary Tree | Set 1

In this post a different O(n) solution is discussed. This solution is simpler than the solutions discussed above and works in O(n) time.

The idea is based on method 3 of check if a binary tree is BST article.

A Tree is BST if following is true for every node x.

1. The largest value in left subtree (of x) is smaller than value of x.
2. The smallest value in right subtree (of x) is greater than value of x.

We traverse tree in bottom up manner. For every traversed node, we return maximum and minimum values in subtree rooted with it. If any node follows above properties and size of

```// C++ program to find largest BST in a
// Binary Tree.
#include <bits/stdc++.h>
using namespace std;

/* A binary tree node has data,
pointer to left child and a
pointer to right child */
struct Node
{
int data;
struct Node* left;
struct Node* right;
};

/* Helper function that allocates a new
node with the given data and NULL left
and right pointers. */
struct Node* newNode(int data)
{
struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;

return(node);
}

// Information to be returned by every
// node in bottom up traversal.
struct Info
{
int sz; // Size of subtree
int max; // Min value in subtree
int min; // Max value in subtree
int ans; // Size of largest BST which
// is subtree of current node
bool isBST; // If subtree is BST
};

// Returns Information about subtree. The
// Information also includes size of largest
// subtree which is a BST.
Info largestBSTBT(Node* root)
{
// Base cases : When tree is empty or it has
// one child.
if (root == NULL)
return {0, INT_MIN, INT_MAX, 0, true};
if (root->left == NULL && root->right == NULL)
return {1, root->data, root->data, 1, true};

// Recur for left subtree and right subtrees
Info l = largestBSTBT(root->left);
Info r = largestBSTBT(root->right);

// Create a return variable and initialize its
// size.
Info ret;
ret.sz = (1 + l.sz + r.sz);

// If whole tree rooted under current root is
// BST.
if (l.isBST && r.isBST && l.max < root->data &&
r.min > root->data)
{
ret.min = min(l.min, min(r.min, root->data));
ret.max = max(r.max, max(l.max, root->data));

// Update answer for tree rooted under
// current 'root'
ret.ans = ret.sz;
ret.isBST = true;

return ret;
}

// If whole tree is not BST, return maximum
// of left and right subtrees
ret.ans = max(l.ans, r.ans);
ret.isBST = false;

return ret;
}

/* Driver program to test above functions*/
int main()
{
/* Let us construct the following Tree
60
/  \
65  70
/
50 */

struct Node *root = newNode(60);
root->left = newNode(65);
root->right = newNode(70);
root->left->left = newNode(50);
printf(" Size of the largest BST is %d\n",
largestBSTBT(root).ans);
return 0;
}

// This code is contributed by Vivek Garg in a
// comment on below set 1.
// www.geeksforgeeks.org/find-the-largest-subtree-in-a-tree-that-is-also-a-bst/
```

Output:

```Size of largest BST is 2
```

Time Complexity : O(n)

This article is contributed by Shubham Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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