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 #include<iostream> using namespace std; // A BST node struct Node { int data; struct Node *left, *right; }; // A utility function to create a new Binary Tree Node struct 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 node void 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 tree void printInorder( struct Node *root) { if (root == NULL) return ; printInorder(root->left); cout << root->data << " " ; printInorder(root->right); } // Driver Program to test above functions int 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 tree import 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 node def 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 tree def printInorder(root): if (root = = None ): return printInorder(root.left) print (root.data, end = " " ) printInorder(root.right) # Driver Program to test above functions if __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 |
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.
https://youtu.be/hx8IADDBqb0?list=PLqM7alHXFySHCXD7r1J0ky9Zg_GBB1dbk
This article is contributed by Bhavana. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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