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 the 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

We have discussed 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++

// 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 {
    // Size of subtree
    int sz;
    // Min value in subtree
    int max;
    // Max value in subtree
    int min;
    // Size of largest BST which
    // is subtree of current node
    int ans;
    // If subtree is BST
    bool isBST;
};

// 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, root->data);
        ret.max = max(r.max, root->data);

        // Update answer for tree rooted under
        // current 'root'
        ret.ans = l.ans + r.ans + 1;
        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/
// This code is updated by Naman Sharma

Java

// Java program to find largest BST in a Binary Tree.

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

    public Node(final int d) { data = d; }
}

class GFG {

    public static void main(String[] args)
    {

        /* Let us construct the following Tree
         60
        /  \
       65  70
      /
     50 */

        final Node node1 = new Node(60);
        node1.left = new Node(65);
        node1.right = new Node(70);
        node1.left.left = new Node(50);

        System.out.print("Size of the largest BST is "
                         + Solution.largestBst(node1)
                         + "\n");
    }
}

class Solution {
    static int MAX = Integer.MAX_VALUE;
    static int MIN = Integer.MIN_VALUE;

    static class nodeInfo {
        int size; // Size of subtree
        int max; // Min value in subtree
        int min; // Max value in subtree
        boolean isBST; // If subtree is BST

        nodeInfo() {}

        nodeInfo(int size, int max, int min, boolean isBST)
        {
            this.size = size;
            this.max = max;
            this.min = min;
            this.isBST = isBST;
        }
    }

    static nodeInfo largestBST(Node root)
    {

        // Base case : When the current subtree is empty or
        // the node corresponds to a null.
        if (root == null) {
            return new nodeInfo(0, Integer.MIN_VALUE,
                                Integer.MAX_VALUE, true);
        }
        // We will here do the postorder traversal since we
        // want our left and right subtrees to be computed
        // first.

        // Recur for left subtree and right subtrees
        nodeInfo left = largestBST(root.left);
        nodeInfo right = largestBST(root.right);

        // Create a new nodeInfo variable to store info
        // about the current node.
        nodeInfo returnInfo = new nodeInfo();

        returnInfo.min = Math.min(left.min, root.data);
        returnInfo.max = Math.max(right.max, root.data);
        returnInfo.isBST = left.isBST && right.isBST
                           && root.data > left.max
                           && root.data < right.min;

        /*
        If suppose the left and right subtrees of the
        current node are BST and the current node value is
        greater than the maximum value in the left subtree
        and also the current node value is smaller that the
        minimum value in the right subtree (Basic conditions
        for the formation of BST) then our whole subtree
        with the root as current root can be consider as a
        BST. Hence the size of the BST will become size of
        left BST subtree + size of right BST subtree +
        1(adding current node).

        Else we will consider the largest of the left BST or
        the  right BST.
        */

        /*We need to find maximum height BST subtree then
        adding the height of left an right subtree will not
        give the maximum height, we need to find max.
        */
        if (returnInfo.isBST)

            // Calculate the size of subtree if
            // this is a BST
            returnInfo.size = left.size + right.size + 1;
        else
            returnInfo.size
                = Math.max(left.size, right.size);

        return returnInfo;
    }

    // Return the size of the largest sub-tree which is also
    // a BST
    static int largestBst(Node root)
    {
        return largestBST(root).size;
    }
}
// This code is contributed by Andrei Sljusar
// This code is updated by Utkarsh Saxena.

Python3

# Python program to find largest
# BST in a Binary Tree.

INT_MIN = -2147483648
INT_MAX = 2147483647

# Helper function that allocates a new
# node with the given data and None left
# and right pointers.
class newNode:

    # Constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# Returns Information about subtree. The
# Information also includes size of largest
# subtree which is a BST
def largestBSTBT(root):

# Base cases : When tree is empty or it has
    # one child.
    if (root == None):
        return 0, INT_MIN, INT_MAX, 0, True
    if (root.left == None and root.right == None) :
        return 1, root.data, root.data, 1, True

    # Recur for left subtree and right subtrees
    l = largestBSTBT(root.left)
    r = largestBSTBT(root.right)

    # Create a return variable and initialize its
    # size.
    ret = [0, 0, 0, 0, 0]
    ret[0] = (1 + l[0] + r[0])

    # If whole tree rooted under current root is
    # BST.
    if (l[4] and r[4] and l[1] <
        root.data and r[2] > root.data) :

        ret[2] = min(l[2], root.data)
        ret[1] = max(r[1], root.data)

        # Update answer for tree rooted under
        # current 'root'
        ret[3] = max(l[3], r[3]) + 1;
        ret[4] = True

        return ret


    # If whole tree is not BST, return maximum
    # of left and right subtrees
    ret[3] = max(l[3], r[3])
    ret[4] = False

    return ret

# Driver Code
if __name__ == '__main__':

    """Let us construct the following Tree
        60
        / \
        65 70
    /
    50 """
    root = newNode(60)
    root.left = newNode(65)
    root.right = newNode(70)
    root.left.left = newNode(50)
    print("Size of the largest BST is",
                    largestBSTBT(root)[3])

# This code is contributed
# Shubham Singh(SHUBHAMSINGH10)
# Naman Sharma (NAMANSHARMA1805)

C#

// C# program to find largest BST in a Binary Tree.

/* A binary tree node has data, pointer to left child and a
 * pointer to right child */
using System;
public

  class Node {
    public

      int data;
    public

      Node left,
    right;

    public Node(int d) { data = d; }
  }

public class GFG {

  public static void Main(String[] args)
  {

    /*
         * Let us construct the following Tree 60 / \ 65 70
         * / 50
         */

    Node node1 = new Node(60);
    node1.left = new Node(65);
    node1.right = new Node(70);
    node1.left.left = new Node(50);

    Console.Write("Size of the largest BST is "
                  + Solution.largestBst(node1) + "\n");
  }
}

public

  class Solution {
    static int MAX = int.MaxValue;
    static int MIN = int.MinValue;

    public

      class nodeInfo {
        public int size; // Size of subtree
        public int max; // Min value in subtree
        public int min; // Max value in subtree
        public bool isBST; // If subtree is BST

        public

          nodeInfo()
        {
        }

        public

          nodeInfo(int size, int max, int min, bool isBST)
        {
          this.size = size;
          this.max = max;
          this.min = min;
          this.isBST = isBST;
        }
      }

    public

      static nodeInfo
      largestBST(Node root)
    {

      // Base cases : When tree is empty or it has one
      // child.
      if (root == null) {
        return new nodeInfo(0, MIN, MAX, true);
      }
      if (root.left == null && root.right == null) {
        return new nodeInfo(1, root.data, root.data,
                            true);
      }

      // Recur for left subtree and right subtrees
      nodeInfo left = largestBST(root.left);
      nodeInfo right = largestBST(root.right);

      // Create a return variable and initialize its size.
      nodeInfo returnInfo = new nodeInfo();

      // If whole tree rooted under current root is BST.
      if (left.isBST && right.isBST
          && left.max < root.data
          && right.min > root.data) {
        returnInfo.min = Math.Min(left.min, root.data);
        returnInfo.max = Math.Max(right.max, root.data);

        // Update answer for tree rooted under current
        // 'root'
        returnInfo.size = Math.Max(left.size, right.size) + 1;
        returnInfo.isBST = true;
        return returnInfo;
      }

      // If whole tree is not BST, return maximum of left
      // and right subtrees
      returnInfo.size = Math.Max(left.size, right.size);
      returnInfo.isBST = false;
      return returnInfo;
    }

    // Return the size of the largest sub-tree which is also
    // a BST
    public

      static int
      largestBst(Node root)
    {
      return largestBST(root).size;
    }
  }

// This code is contributed by Rajput-Ji
//This code is updated by Naman Sharma

Javascript

<script>
// javascript program to find largest BST in a Binary Tree.

/* A binary tree node has data, pointer to left child and a pointer to right child */
class Node {
    constructor(val) {
        this.data = val;
        this.left = null;
        this.right = null;
    }
}

var MAX = Number.MAX_VALUE;
    var MIN = Number.MIN_VALUE;

     class nodeInfo {

        constructor(size , max , min,  isBST) {
            this.size = size;
            this.max = max;
            this.min = min;
            this.isBST = isBST;
        }
    }

     function largestBST(root) {

        // Base cases : When tree is empty or it has one child.
        if (root == null) {
            return new nodeInfo(0, MIN, MAX, true);
        }
        if (root.left == null && root.right == null) {
            return new nodeInfo(1, root.data, root.data, true);
        }

        // Recur for left subtree and right subtrees
        var left = largestBST(root.left);
        var right = largestBST(root.right);

        // Create a return variable and initialize its size.
        var returnInfo = new nodeInfo();

        // If whole tree rooted under current root is BST.
        if (left.isBST && right.isBST && left.max < root.data && right.min > root.data) {
            returnInfo.min = Math.min(left.min, root.data);
            returnInfo.max = Math.max(right.max, root.data);

            // Update answer for tree rooted under current 'root'
            returnInfo.size = math.max(left.size, right.size) + 1;
            returnInfo.isBST = true;
            return returnInfo;
        }

        // If whole tree is not BST, return maximum of left and right subtrees
        returnInfo.size = Math.max(left.size, right.size);
        returnInfo.isBST = false;
        return returnInfo;
    }

    // Return the size of the largest sub-tree which is also a BST
    function largestBst(root) {
        return largestBST(root).size;
}

        /*
         * Let us construct the following Tree
         60
        / \
       65 70
      /
      50
         */

        var node1 = new Node(60);
        node1.left = new Node(65);
        node1.right = new Node(70);
        node1.left.left = new Node(50);

        document.write("Size of the largest BST is " + largestBst(node1) + "\n");

// This code is contributed by Rajput-Ji and Naman Sharma
</script>

Output

 Size of the largest BST is 2

Time Complexity : O(n)

Space complexity: O(n) For call stack since using recursion

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