Transform a BST to greater sum tree

• Difficulty Level : Medium
• Last Updated : 22 Jun, 2021

Given a BST, transform it into a greater sum tree where each node contains sum of all nodes greater than that node. We strongly recommend to minimize the browser and try this yourself first.
Method 1 (Naïve):
This method doesn’t require the tree to be a BST. Following are the steps.
1. Traverse node by node(Inorder, preorder, etc.)
2. For each node find all the nodes greater than that of the current node, sum the values. Store all these sums.
3. Replace each node value with their corresponding sum by traversing in the same order as in Step 1.
This takes O(n^2) Time Complexity.

Method 2 (Using only one traversal)
By leveraging the fact that the tree is a BST, we can find an O(n) solution. The idea is to traverse BST in reverse inorder. Reverse inorder traversal of a BST gives us keys in decreasing order. Before visiting a node, we visit all greater nodes of that node. While traversing we keep track of the sum of keys which is the sum of all the keys greater than the key of the current node.

C++

 // C++ program to transform a BST to sum tree#includeusing namespace std; // A BST nodestruct Node{    int data;    struct Node *left, *right;}; // A utility function to create a new Binary Tree Nodestruct Node *newNode(int item){    struct Node *temp =  new Node;    temp->data = item;    temp->left = temp->right = NULL;    return temp;} // Recursive function to transform a BST to sum tree.// This function traverses the tree in reverse inorder so// that we have visited all greater key nodes of the currently// visited nodevoid transformTreeUtil(struct Node *root, int *sum){   // Base case   if (root == NULL)  return;    // Recur for right subtree   transformTreeUtil(root->right, sum);    // Update sum   *sum = *sum + root->data;    // Store old sum in current node   root->data = *sum - root->data;    // Recur for left subtree   transformTreeUtil(root->left, sum);} // A wrapper over transformTreeUtil()void transformTree(struct Node *root){    int sum = 0; // Initialize sum    transformTreeUtil(root, &sum);} // A utility function to print indorder traversal of a// binary treevoid printInorder(struct Node *root){    if (root == NULL) return;     printInorder(root->left);    cout << root->data << " ";    printInorder(root->right);} // Driver Program to test above functionsint main(){    struct Node *root = newNode(11);    root->left = newNode(2);    root->right = newNode(29);    root->left->left = newNode(1);    root->left->right = newNode(7);    root->right->left = newNode(15);    root->right->right = newNode(40);    root->right->right->left = newNode(35);     cout << "Inorder Traversal of given tree\n";    printInorder(root);     transformTree(root);     cout << "\n\nInorder Traversal of transformed tree\n";    printInorder(root);     return 0;}

Java

 // Java program to transform a BST to sum treeimport java.io.*;class Node{  int data;  Node left, right;   // A utility function to create a new Binary Tree Node  Node(int item)  {    data = item;    left = right = null;  }} class GFG{   static int sum = 0;  static Node Root;   // Recursive function to transform a BST to sum tree.  // This function traverses the tree in reverse inorder so  // that we have visited all greater key nodes of the currently  // visited node  static void transformTreeUtil(Node root)  {     // Base case    if (root == null)       return;     // Recur for right subtree    transformTreeUtil(root.right);     // Update sum    sum = sum + root.data;     // Store old sum in current node    root.data = sum - root.data;     // Recur for left subtree    transformTreeUtil(root.left);  }   // A wrapper over transformTreeUtil()  static void transformTree(Node root)  {     transformTreeUtil(root);  }   // A utility function to print indorder traversal of a  // binary tree  static void printInorder(Node root)  {    if (root == null)      return;    printInorder(root.left);    System.out.print(root.data + " ");    printInorder(root.right);  }   // Driver Program to test above functions  public static void main (String[] args) {     GFG.Root = new Node(11);    GFG.Root.left = new Node(2);    GFG.Root.right = new Node(29);    GFG.Root.left.left = new Node(1);    GFG.Root.left.right = new Node(7);    GFG.Root.right.left = new Node(15);    GFG.Root.right.right = new Node(40);    GFG.Root.right.right.left = new Node(35);     System.out.println("Inorder Traversal of given tree");    printInorder(Root);     transformTree(Root);    System.out.println("\n\nInorder Traversal of transformed tree");    printInorder(Root);  }} // This code is contributed by avanitrachhadiya2155

Python3

 # Python3 program to transform a BST to sum tree class Node:    def __init__(self, x):        self.data = x        self.left = None        self.right = None # Recursive function to transform a BST to sum tree.# This function traverses the tree in reverse inorder so# that we have visited all greater key nodes of the currently# visited nodedef transformTreeUtil(root):      # Base case   if (root == None):        return    # Recur for right subtree   transformTreeUtil(root.right)    # Update sum   global sum   sum = sum + root.data    # Store old sum in current node   root.data = sum - root.data    # Recur for left subtree   transformTreeUtil(root.left) # A wrapper over transformTreeUtil()def transformTree(root):     # sum = 0 #Initialize sum    transformTreeUtil(root) # A utility function to prindorder traversal of a# binary treedef printInorder(root):    if (root == None):        return     printInorder(root.left)    print(root.data, end = " ")    printInorder(root.right) # Driver Program to test above functionsif __name__ == '__main__':     sum=0    root = Node(11)    root.left = Node(2)    root.right = Node(29)    root.left.left = Node(1)    root.left.right = Node(7)    root.right.left = Node(15)    root.right.right = Node(40)    root.right.right.left = Node(35)     print("Inorder Traversal of given tree")    printInorder(root)     transformTree(root)     print("\nInorder Traversal of transformed tree")    printInorder(root)     # This code is contributed by mohit kumar 29

C#

 // C# program to transform a BST to sum treeusing System; public class Node{  public int data;  public Node left, right;   // A utility function to create a new Binary Tree Node  public Node(int item)  {    data = item;    left = right = null;  }} public class GFG{   static int sum = 0;  static Node Root;   // Recursive function to transform a BST to sum tree.  // This function traverses the tree in reverse inorder so  // that we have visited all greater key nodes of the currently  // visited node  static void transformTreeUtil(Node root)  {     // Base case    if (root == null)      return;     // Recur for right subtree    transformTreeUtil(root.right);     // Update sum    sum = sum + root.data;     // Store old sum in current node    root.data = sum - root.data;     // Recur for left subtree    transformTreeUtil(root.left);  }   // A wrapper over transformTreeUtil()  static void transformTree(Node root)  {     transformTreeUtil(root);  }   // A utility function to print indorder traversal of a  // binary tree  static void printInorder(Node root)  {    if (root == null)      return;    printInorder(root.left);    Console.Write(root.data + " ");    printInorder(root.right);  }   // Driver Program to test above functions   static public void Main (){    GFG.Root = new Node(11);    GFG.Root.left = new Node(2);    GFG.Root.right = new Node(29);    GFG.Root.left.left = new Node(1);    GFG.Root.left.right = new Node(7);    GFG.Root.right.left = new Node(15);    GFG.Root.right.right = new Node(40);    GFG.Root.right.right.left = new Node(35);     Console.WriteLine("Inorder Traversal of given tree");    printInorder(Root);     transformTree(Root);    Console.WriteLine("\n\nInorder Traversal of transformed tree");    printInorder(Root);  }} // This code is contributed by ab2127

Javascript



Output:

Inorder Traversal of given tree
1 2 7 11 15 29 35 40

Inorder Traversal of transformed tree
139 137 130 119 104 75 40 0

The time complexity of this method is O(n) as it does a simple traversal of the tree.