# Maximum sub-tree sum in a Binary Tree such that the sub-tree is also a BST

Given a binary tree, the task is to print the maximum sum of nodes of a sub-tree which is also a Binary Search Tree.
Examples:

Input :
7
/  \
12    2
/  \    \
11  13    5
/         / \
2         1   38

Output:44
BST rooted under node 5 has the maximum sum
5
/ \
1   38

Input:
5
/  \
9    2
/      \
6        3
/ \
8   7

Output: 8
Here each leaf node represents a binary search tree
also a BST with sum 5 exists
2
\
3
But the leaf node 8 has the maximum sum.

Approach: We traverse the tree in bottom-up manner. For every traversed node, we store the information of maximum and minimum of that subtree, a variable isBST to store if it is a BST, variable currmax to store the maximum sum of BST found till now, and a variable sum to store the sum of Left and Right subtree(which is also a BST) rooted under the current node.
Below is the implementation of the above approach:

 // C++ implementation of the approach #include using namespace std;   // Binary tree node struct Node {     struct Node* left;     struct Node* right;     int data;       Node(int data)     {         this->data = data;         this->left = NULL;         this->right = NULL;     } };   // Information stored in every // node during bottom up traversal struct Info {       // Max Value in the subtree     int max;       // Min value in the subtree     int min;       // If subtree is BST     bool isBST;       // Sum of the nodes of the sub-tree     // rooted under the current node     int sum;       // Max sum of BST found till now     int currmax; };   // Returns information about subtree such as // subtree with maximum sum which is also a BST Info MaxSumBSTUtil(struct Node* root, int& maxsum) {     // Base case     if (root == NULL)         return { INT_MIN, INT_MAX, true, 0, 0 };       // If current node is a leaf node then     // return from the function and store     // information about the leaf node     if (root->left == NULL && root->right == NULL) {         maxsum = max(maxsum, root->data);         return { root->data, root->data, true, root->data, maxsum };     }       // Store information about the left subtree     Info L = MaxSumBSTUtil(root->left, maxsum);       // Store information about the right subtree     Info R = MaxSumBSTUtil(root->right, maxsum);       Info BST;       // If the subtree rooted under the current node     // is a BST     if (L.isBST && R.isBST && L.max < root->data && R.min > root->data) {           BST.max = max(root->data, max(L.max, R.max));         BST.min = min(root->data, min(L.min, R.min));           maxsum = max(maxsum, R.sum + root->data + L.sum);         BST.sum = R.sum + root->data + L.sum;           // Update the current maximum sum         BST.currmax = maxsum;           BST.isBST = true;         return BST;     }       // If the whole tree is not a BST then     // update the current maximum sum     BST.isBST = false;     BST.currmax = maxsum;     BST.sum = R.sum + root->data + L.sum;       return BST; }   // Function to return the maximum // sum subtree which is also a BST int MaxSumBST(struct Node* root) {     int maxsum = INT_MIN;     return MaxSumBSTUtil(root, maxsum).currmax; }   // Driver code int main() {     struct Node* root = new Node(5);     root->left = new Node(14);     root->right = new Node(3);     root->left->left = new Node(6);     root->right->right = new Node(7);     root->left->left->left = new Node(9);     root->left->left->right = new Node(1);       cout << MaxSumBST(root);       return 0; }

 // Java implementation of the approach class GFG {       // Binary tree node static class Node {     Node left;     Node right;     int data;       Node(int data)     {         this.data = data;         this.left = null;         this.right = null;     } };   // Information stored in every // node during bottom up traversal static class Info {       // Max Value in the subtree     int max;       // Min value in the subtree     int min;       // If subtree is BST     boolean isBST;       // Sum of the nodes of the sub-tree     // rooted under the current node     int sum;       // Max sum of BST found till now     int currmax;           Info(int m,int mi,boolean is,int su,int cur)     {         max = m;         min = mi;         isBST = is;         sum = su;         currmax = cur;     }     Info(){} };   static class INT {     int a; }   // Returns information about subtree such as // subtree with the maximum sum which is also a BST static Info MaxSumBSTUtil( Node root, INT maxsum) {     // Base case     if (root == null)         return new Info( Integer.MIN_VALUE,                         Integer.MAX_VALUE, true, 0, 0 );       // If current node is a leaf node then     // return from the function and store     // information about the leaf node     if (root.left == null && root.right == null)     {         maxsum.a = Math.max(maxsum.a, root.data);         return new Info( root.data, root.data,                         true, root.data, maxsum.a );     }       // Store information about the left subtree     Info L = MaxSumBSTUtil(root.left, maxsum);       // Store information about the right subtree     Info R = MaxSumBSTUtil(root.right, maxsum);       Info BST=new Info();       // If the subtree rooted under the current node     // is a BST     if (L.isBST && R.isBST && L.max < root.data &&                                R.min > root.data)     {           BST.max = Math.max(root.data, Math.max(L.max, R.max));         BST.min = Math.min(root.data, Math.min(L.min, R.min));           maxsum.a = Math.max(maxsum.a, R.sum + root.data + L.sum);         BST.sum = R.sum + root.data + L.sum;           // Update the current maximum sum         BST.currmax = maxsum.a;           BST.isBST = true;         return BST;     }       // If the whole tree is not a BST then     // update the current maximum sum     BST.isBST = false;     BST.currmax = maxsum.a;     BST.sum = R.sum + root.data + L.sum;       return BST; }   // Function to return the maximum // sum subtree which is also a BST static int MaxSumBST( Node root) {     INT maxsum = new INT();     maxsum.a = Integer.MIN_VALUE;     return MaxSumBSTUtil(root, maxsum).currmax; }   // Driver code public static void main(String args[]) {     Node root = new Node(5);     root.left = new Node(14);     root.right = new Node(3);     root.left.left = new Node(6);     root.right.right = new Node(7);     root.left.left.left = new Node(9);     root.left.left.right = new Node(1);       System.out.println( MaxSumBST(root)); } }   // This code is contributed by Arnab Kundu

 # Python3 implementation of # the above approach from sys import maxsize as INT_MAX INT_MIN = -INT_MAX   # Binary tree node class Node:         def __init__(self, data):         self.data = data         self.left = None         self.right = None   # Information stored in every # node during bottom up traversal class Info:         def __init__(self, _max, _min,                  isBST, _sum, currmax):                 # Max Value in the subtree         self.max = _max           # Min value in the subtree         self.min = _min           # If subtree is BST         self.isBST = isBST           # Sum of the nodes of the sub-tree         # rooted under the current node         self.sum = _sum           # Max sum of BST found till now         self.currmax = currmax   # Returns information about # subtree such as subtree # with maximum sum which # is also a BST def MaxSumBSTUtil(root: Node) -> Info:     global maxsum       # Base case     if (root is None):         return Info(INT_MIN, INT_MAX,                     True, 0, 0)       # If current node is a     # leaf node then return     # from the function and store     # information about the leaf node     if (root.left is None and         root.right is None):         maxsum = max(maxsum,                      root.data)         return Info(root.data, root.data,                     True, root.data, maxsum)       # Store information about     # the left subtree     L = MaxSumBSTUtil(root.left)       # Store information about     # the right subtree     R = MaxSumBSTUtil(root.right)       BST = Info       # If the subtree rooted under     # the current node is a BST     if (L.isBST and R.isBST and         L.max < root.data and         R.min > root.data):           BST.max = max(root.data,                   max(L.max, R.max))         BST.min = min(root.data,                   min(L.min, R.min))           maxsum = max(maxsum, R.sum +                      root.data + L.sum)         BST.sum = R.sum + root.data + L.sum           # Update the current maximum sum         BST.currmax = maxsum           BST.isBST = True         return BST       # If the whole tree is not     # a BST then update the     # current maximum sum     BST.isBST = False     BST.currmax = maxsum     BST.sum = R.sum + root.data + L.sum       return BST   # Function to return the maximum # sum subtree which is also a BST def MaxSumBST(root: Node) -> int:     global maxsum     return MaxSumBSTUtil(root).currmax   # Driver code if __name__ == "__main__":       root = Node(5)     root.left = Node(14)     root.right = Node(3)     root.left.left = Node(6)     root.right.right = Node(7)     root.left.left.left = Node(9)     root.left.left.right = Node(1)       maxsum = INT_MIN     print(MaxSumBST(root))   # This code is contributed by sanjeev2552

 // C# implementation of the approach using System;   class GFG {       // Binary tree node public class Node {     public Node left;     public Node right;     public int data;       public Node(int data)     {         this.data = data;         this.left = null;         this.right = null;     } };   // Information stored in every // node during bottom up traversal public class Info {       // Max Value in the subtree     public int max;       // Min value in the subtree     public int min;       // If subtree is BST     public bool isBST;       // Sum of the nodes of the sub-tree     // rooted under the current node     public int sum;       // Max sum of BST found till now     public int currmax;           public Info(int m,int mi,bool s,int su,int cur)     {         max = m;         min = mi;         isBST = s;         sum = su;         currmax = cur;     }     public Info(){} };   public class INT {     public int a; }   // Returns information about subtree such as // subtree with the maximum sum which is also a BST static Info MaxSumBSTUtil( Node root, INT maxsum) {     // Base case     if (root == null)         return new Info( int.MinValue,                         int.MaxValue, true, 0, 0 );       // If current node is a leaf node then     // return from the function and store     // information about the leaf node     if (root.left == null && root.right == null)     {         maxsum.a = Math.Max(maxsum.a, root.data);         return new Info( root.data, root.data,                         true, root.data, maxsum.a );     }       // Store information about the left subtree     Info L = MaxSumBSTUtil(root.left, maxsum);       // Store information about the right subtree     Info R = MaxSumBSTUtil(root.right, maxsum);       Info BST = new Info();       // If the subtree rooted under the current node     // is a BST     if (L.isBST && R.isBST && L.max < root.data &&                             R.min > root.data)     {           BST.max = Math.Max(root.data, Math.Max(L.max, R.max));         BST.min = Math.Min(root.data, Math.Min(L.min, R.min));           maxsum.a = Math.Max(maxsum.a, R.sum + root.data + L.sum);         BST.sum = R.sum + root.data + L.sum;           // Update the current maximum sum         BST.currmax = maxsum.a;           BST.isBST = true;         return BST;     }       // If the whole tree is not a BST then     // update the current maximum sum     BST.isBST = false;     BST.currmax = maxsum.a;     BST.sum = R.sum + root.data + L.sum;       return BST; }   // Function to return the maximum // sum subtree which is also a BST static int MaxSumBST( Node root) {     INT maxsum = new INT();     maxsum.a = int.MinValue;     return MaxSumBSTUtil(root, maxsum).currmax; }   // Driver code public static void Main(String []args) {     Node root = new Node(5);     root.left = new Node(14);     root.right = new Node(3);     root.left.left = new Node(6);     root.right.right = new Node(7);     root.left.left.left = new Node(9);     root.left.left.right = new Node(1);       Console.WriteLine( MaxSumBST(root)); } }   // This code has been contributed by 29AjayKumar

Output:
10

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