Add all greater values to every node in a given BST

2.5

Given a Binary Search Tree (BST), modify it so that all greater values in the given BST are added to every node. For example, consider the following BST.

              50
           /      \
         30        70
        /   \      /  \
      20    40    60   80 

The above tree should be modified to following 

              260
           /      \
         330        150
        /   \       /  \
      350   300    210   80

A simple method for solving this is to find sum of all greater values for every node. This method would take O(n^2) time.
We can do it using a single traversal. The idea is to use following BST property. If we do reverse Inorder traversal of BST, we get all nodes in decreasing order. We do reverse Inorder traversal and keep track of the sum of all nodes visited so far, we add this sum to every node.

C

// C program to add all greater values in every node of BST
#include<stdio.h>
#include<stdlib.h>

struct Node
{
    int data;
    struct Node *left, *right;
};

// A utility function to create a new BST node
struct Node *newNode(int item)
{
    struct Node *temp =  (struct Node *)malloc(sizeof(struct Node));
    temp->data = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Recursive function to add all greater values in every node
void modifyBSTUtil(struct Node *root, int *sum)
{
    // Base Case
    if (root == NULL)  return;

    // Recur for right subtree
    modifyBSTUtil(root->right, sum);

    // Now *sum has sum of nodes in right subtree, add
    // root->data to sum and update root->data
    *sum = *sum + root->data;
    root->data = *sum;

    // Recur for left subtree
    modifyBSTUtil(root->left, sum);
}

// A wrapper over modifyBSTUtil()
void modifyBST(struct Node *root)
{
    int sum = 0;
    modifyBSTUtil(root, &sum);
}

// A utility function to do inorder traversal of BST
void inorder(struct Node *root)
{
    if (root != NULL)
    {
        inorder(root->left);
        printf("%d ", root->data);
        inorder(root->right);
    }
}

/* A utility function to insert a new node with given data in BST */
struct Node* insert(struct Node* node, int data)
{
    /* If the tree is empty, return a new node */
    if (node == NULL) return newNode(data);

    /* Otherwise, recur down the tree */
    if (data <= node->data)
        node->left  = insert(node->left, data);
    else
        node->right = insert(node->right, data);

    /* return the (unchanged) node pointer */
    return node;
}

// Driver Program to test above functions
int main()
{
    /* Let us create following BST
              50
           /     \
          30      70
         /  \    /  \
       20   40  60   80 */
    struct Node *root = NULL;
    root = insert(root, 50);
    insert(root, 30);
    insert(root, 20);
    insert(root, 40);
    insert(root, 70);
    insert(root, 60);
    insert(root, 80);

    modifyBST(root);

    // print inoder tarversal of the modified BST
    inorder(root);

    return 0;
}

Java

// Java code to add all greater values to 
// every node in a given BST

// A binary tree node
class Node {

    int data;
    Node left, right;

    Node(int d)
    {
        data = d;
        left = right = null;
    }
}

class BinarySearchTree {

    // Root of BST
    Node root;

    // Constructor
    BinarySearchTree()
    {
        root = null;
    }

    // Inorder traversal of the tree
    void inorder()
    {
        inorderUtil(this.root);
    }

    // Utility function for inorder traversal of
    // the tree
    void inorderUtil(Node node)
    {
        if (node == null)
            return;

        inorderUtil(node.left);
        System.out.print(node.data + " ");
        inorderUtil(node.right);
    }

    // adding new node 
    public void insert(int data)
    {
        this.root = this.insertRec(this.root, data);
    }
    
    /* A utility function to insert a new node with 
    given data in BST */
    Node insertRec(Node node, int data)
    {   
        /* If the tree is empty, return a new node */
        if (node == null) {
            this.root = new Node(data);
            return this.root;
        }

        /* Otherwise, recur down the tree */
        if (data <= node.data) {
            node.left = this.insertRec(node.left, data);
        } else {
            node.right = this.insertRec(node.right, data);
        }
        return node;
    }

    // This class initialises the value of sum to 0
    public class Sum {
        int sum = 0;
    }

    // Recursive function to add all greater values in
    // every node
    void modifyBSTUtil(Node node, Sum S)
    {
        // Base Case
        if (node == null)
            return;
            
        // Recur for right subtree    
        this.modifyBSTUtil(node.right, S); 
        
        // Now *sum has sum of nodes in right subtree, add
        // root->data to sum and update root->data
        S.sum = S.sum + node.data;
        node.data = S.sum;
        
        // Recur for left subtree
        this.modifyBSTUtil(node.left, S); 
    }

    // A wrapper over modifyBSTUtil()
    void modifyBST(Node node)
    {
        Sum S = new Sum();
        this.modifyBSTUtil(node, S);
    }

    // Driver Function
    public static void main(String[] args)
    {
        BinarySearchTree tree = new BinarySearchTree();
        
          /* Let us create following BST
              50
           /     \
          30      70
         /  \    /  \
       20   40  60   80 */
       
        tree.insert(50);
        tree.insert(30);
        tree.insert(20);
        tree.insert(40);
        tree.insert(70);
        tree.insert(60);
        tree.insert(80);

        tree.modifyBST(tree.root);
        
        // print inoder tarversal of the modified BST
        tree.inorder();
    }
}

// This code is contributed by Kamal Rawal


Output
350 330 300 260 210 150 80

Time Complexity: O(n) where n is number of nodes in the given BST.

As a side note, we can also use reverse Inorder traversal to find kth largest element in a BST.

Asked in: Amazon

This article is contributed by Chandra Prakash. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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