Add all greater values to every node in a given BST

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 the sum of all greater values for every node. This method would take O(n^2) time.

The method discussed in this article uses the technique of reverse in-order tree traversal of BST which optimises the problem to be solved in a single traversal.
Approach: In this problem as we could notice that the largest node would remain the same. The value of 2nd largest node = value of largest + value of second largest node. Similarly, the value of nth largest node will be the sum of the n-th node and value of (n-1)th largest node after modification. So if we traverse the tree in descending order and simultaneously update the sum value at every step while adding the value to the root node, the problem would be solved.
So to traverse the BST in descending order we use reverse in-order traversal of BST. This takes a global variable sum which is updated at every node and once the root node is reached it is added to the value of root node and value of the root node is updated.

C++

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// C++ program to add all greater
// values in every node of BST
#include <bits/stdc++.h>
using namespace std;
  
class Node {
public:
    int data;
    Node *left, *right;
};
  
// A utility function to create
// a new BST node
Node* newNode(int item)
{
    Node* temp = new Node();
    temp->data = item;
    temp->left = temp->right = NULL;
    return temp;
}
  
// Recursive function to add all
// greater values in every node
void modifyBSTUtil(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(Node* root)
{
    int sum = 0;
    modifyBSTUtil(root, &sum);
}
  
// A utility function to do
// inorder traversal of BST
void inorder(Node* root)
{
    if (root != NULL) {
        inorder(root->left);
        cout << root->data << " ";
        inorder(root->right);
    }
}
  
/* A utility function to insert 
a new node with given data in BST */
Node* insert(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 code
int main()
{
    /* Let us create following BST 
            50 
        / \ 
        30 70 
        / \ / \ 
    20 40 60 80 */
    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;
}
  
// This code is contributed by rathbhupendra

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C

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// 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;
}

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Java

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// 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

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Python3

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# Python3 program to add all greater values 
# in every node of BST 
  
# A utility function to create a
# new BST node 
class newNode: 
  
    # Constructor to create a new node 
    def __init__(self, data): 
        self.data = data 
        self.left = None
        self.right = None
  
# Recursive function to add all greater
# values in every node 
def modifyBSTUtil(root, Sum):
      
    # Base Case 
    if root == None:
        return
  
    # Recur for right subtree 
    modifyBSTUtil(root.right, Sum
  
    # Now Sum[0] has sum of nodes in right 
    # subtree, add root.data to sum and
    # update root.data 
    Sum[0] = Sum[0] + root.data 
    root.data = Sum[0]
  
    # Recur for left subtree 
    modifyBSTUtil(root.left, Sum)
  
# A wrapper over modifyBSTUtil() 
def modifyBST(root):
    Sum = [0]
    modifyBSTUtil(root, Sum)
  
# A utility function to do inorder
# traversal of BST 
def inorder(root):
    if root != None:
        inorder(root.left)
        print(root.data, end =" "
        inorder(root.right)
  
# A utility function to insert a new node 
# with given data in BST
def insert(node, data):
      
    # If the tree is empty, return a new node
    if node == None:
        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 Code
if __name__ == '__main__':
      
    # Let us create following BST 
    # 50 
    #     /     \ 
    # 30     70 
    #     / \ / \ 
    # 20 40 60 80
    root = None
    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)
      
# This code is contributed by PranchalK

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C#

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using System;
  
// C# code to add all greater values to
// every node in a given BST
  
// A binary tree node
public class Node {
  
    public int data;
    public Node left, right;
  
    public Node(int d)
    {
        data = d;
        left = right = null;
    }
}
  
public class BinarySearchTree {
  
    // Root of BST
    public Node root;
  
    // Constructor
    public BinarySearchTree()
    {
        root = null;
    }
  
    // Inorder traversal of the tree
    public virtual void inorder()
    {
        inorderUtil(this.root);
    }
  
    // Utility function for inorder traversal of
    // the tree
    public virtual void inorderUtil(Node node)
    {
        if (node == null) {
            return;
        }
  
        inorderUtil(node.left);
        Console.Write(node.data + " ");
        inorderUtil(node.right);
    }
  
    // adding new node
    public virtual void insert(int data)
    {
        this.root = this.insertRec(this.root, data);
    }
  
    /* A utility function to insert a new node with  
    given data in BST */
    public virtual 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 {
        private readonly BinarySearchTree outerInstance;
  
        public Sum(BinarySearchTree outerInstance)
        {
            this.outerInstance = outerInstance;
        }
  
        public int sum = 0;
    }
  
    // Recursive function to add all greater values in
    // every node
    public virtual 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()
    public virtual void modifyBST(Node node)
    {
        Sum S = new Sum(this);
        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 Shrikant13

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Output:

350 330 300 260 210 150 80

Complexity Analysis:



  • Time Complexity: O(n).
    As this problem uses an in-order tree traversal technique
  • Auxiliary Space: O(1).
    As no data structure has been used for storing values.

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

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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