Convert a Binary Tree such that every node stores the sum of all nodes in its right subtree

Given a binary tree, change the value in each node to sum of all the values in the nodes in the right subtree including its own.

Examples:

Input : 
     1
   /   \
 2      3
Output : 
    4
  /   \
 2     3

Input : 
       1
      / \
     2   3
    / \   \
   4   5   6
Output :
       10
      / \
     7   9
    / \   \
   4   5   6


Approach : The idea is to traverse the given binary tree in bottom up manner. Recursively compute the sum of nodes in right and left subtrees. Accumulate sum of nodes in the right subtree to the current node and return sum of nodes under current subtree.

Below is the implementation of above approach.

C++

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// C++ program to store sum of nodes in
// right subtree in every node
#include <bits/stdc++.h>
using namespace std;
  
// Node of tree
struct Node {
    int data;
    Node *left, *right;
};
  
// Function to create a new node
struct Node* createNode(int item)
{
    Node* temp = new Node;
    temp->data = item;
    temp->left = NULL;
    temp->right = NULL;
  
    return temp;
}
  
// Function to build new tree with 
// all nodes having the sum of all 
// nodes in its right subtree
int updateBTree(Node* root)
{
    // Base cases
    if (!root)
        return 0;
    if (root->left == NULL && root->right == NULL)
        return root->data;
  
    // Update right and left subtrees
    int rightsum = updateBTree(root->right);
    int leftsum = updateBTree(root->left);
  
    // Add rightsum to current node
    root->data += rightsum;
  
    // Return sum of values under root
    return root->data + leftsum;
}
  
// Function to traverse tree in inorder way
void inorder(struct Node* node)
{
    if (node == NULL)
        return;
    inorder(node->left);
    cout << node->data << " ";
    inorder(node->right);
}
  
// Driver code
int main()
{
    /* Let us construct a binary tree
            1
           / \
          2   3
         / \   \
        4   5   6       */
    struct Node* root = NULL;
    root = createNode(1);
    root->left = createNode(2);
    root->right = createNode(3);
    root->left->left = createNode(4);
    root->left->right = createNode(5);
    root->right->right = createNode(6);
  
    // new tree construction
    updateBTree(root);
  
    cout << "Inorder traversal of the modified tree is \n";
    inorder(root);
  
    return 0;
}

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Java

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// Java program to store sum of nodes in 
// right subtree in every node 
class GFG
{
  
// Node of tree 
static class Node 
    int data; 
    Node left, right; 
}; 
  
// Function to create a new node 
static Node createNode(int item) 
    Node temp = new Node(); 
    temp.data = item; 
    temp.left = null
    temp.right = null
  
    return temp; 
  
// Function to build new tree with 
// all nodes having the sum of all 
// nodes in its right subtree 
static int updateBTree(Node root) 
    // Base cases 
    if (root == null
        return 0
    if (root.left == null && root.right == null
        return root.data; 
  
    // Update right and left subtrees 
    int rightsum = updateBTree(root.right); 
    int leftsum = updateBTree(root.left); 
  
    // Add rightsum to current node 
    root.data += rightsum; 
  
    // Return sum of values under root 
    return root.data + leftsum; 
  
// Function to traverse tree in inorder way 
static void inorder( Node node) 
    if (node == null
        return
    inorder(node.left); 
    System.out.print( node.data + " "); 
    inorder(node.right); 
  
// Driver code 
public static void main(String args[])
    /* Let us construct a binary tree 
        
        / \ 
        2 3 
        / \ \ 
        4 5 6 */
    Node root = null
    root = createNode(1); 
    root.left = createNode(2); 
    root.right = createNode(3); 
    root.left.left = createNode(4); 
    root.left.right = createNode(5); 
    root.right.right = createNode(6); 
  
    // new tree conion 
    updateBTree(root); 
  
    System.out.print( "Inorder traversal of the modified tree is \n"); 
    inorder(root); 
}
  
// This code is contributed by Arnab Kundu

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Python3

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# Program to convert expression tree 
# from prefix expression
  
# Helper function that allocates a new 
# node with the given data and None 
# left and right poers.                                 
class createNode: 
  
    # Conto create a new node 
    def __init__(self, key): 
        self.data = key
        self.left = None
        self.right = None
          
# Function to build new tree with 
# all nodes having the sum of all 
# nodes in its right subtree
def updateBTree( root):
  
    # Base cases
    if (not root):
        return 0
    if (root.left == None and
        root.right == None):
        return root.data
  
    # Update right and left subtrees
    rightsum = updateBTree(root.right)
    leftsum = updateBTree(root.left)
  
    # Add rightsum to current node
    root.data += rightsum
  
    # Return sum of values under root
    return root.data + leftsum
  
# Function to traverse tree in inorder way
def inorder(node):
  
    if (node == None):
        return
    inorder(node.left)
    print(node.data, end = " ")
    inorder(node.right)
  
# Driver Code 
if __name__ == '__main__':
      
    """ Let us convert binary tree
        1
    / \
    2 3
    / \ \
    4 5 6 """
    root = None
    root = createNode(1)
    root.left = createNode(2)
    root.right = createNode(3)
    root.left.left = createNode(4)
    root.left.right = createNode(5)
    root.right.right = createNode(6)
  
    # new tree construction
    updateBTree(root)
  
    print("Inorder traversal of the"
          "modified tree is")
    inorder(root)
  
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

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C#

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// C# program to store sum of nodes in 
// right subtree in every node 
using System;
      
class GFG
{
  
// Node of tree 
public class Node 
    public int data; 
    public Node left, right; 
}; 
  
// Function to create a new node 
static Node createNode(int item) 
    Node temp = new Node(); 
    temp.data = item; 
    temp.left = null
    temp.right = null
  
    return temp; 
  
// Function to build new tree with 
// all nodes having the sum of all 
// nodes in its right subtree 
static int updateBTree(Node root) 
    // Base cases 
    if (root == null
        return 0; 
    if (root.left == null && root.right == null
        return root.data; 
  
    // Update right and left subtrees 
    int rightsum = updateBTree(root.right); 
    int leftsum = updateBTree(root.left); 
  
    // Add rightsum to current node 
    root.data += rightsum; 
  
    // Return sum of values under root 
    return root.data + leftsum; 
  
// Function to traverse tree in inorder way 
static void inorder( Node node) 
    if (node == null
        return
    inorder(node.left); 
    Console.Write( node.data + " "); 
    inorder(node.right); 
  
// Driver code 
public static void Main(String[] args)
    /* Let us construct a binary tree 
        
        / \ 
        2 3 
        / \ \ 
        4 5 6 */
    Node root = null
    root = createNode(1); 
    root.left = createNode(2); 
    root.right = createNode(3); 
    root.left.left = createNode(4); 
    root.left.right = createNode(5); 
    root.right.right = createNode(6); 
  
    // new tree conion 
    updateBTree(root); 
  
    Console.Write( "Inorder traversal of the modified tree is \n"); 
    inorder(root); 
}
}
  
// This code contributed by Rajput-Ji

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Output:

Inorder traversal of the modified tree is 
4 7 5 10 9 6

Time Complexity: O(n)



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