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++
// 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; } |
chevron_right
filter_none
Java
// 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 1 / \ 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 |
chevron_right
filter_none
Python3
# 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) |
chevron_right
filter_none
C#
// 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 1 / \ 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 |
chevron_right
filter_none
Output:
Inorder traversal of the modified tree is 4 7 5 10 9 6
Time Complexity: O(n)
Recommended Posts:
- Change a Binary Tree so that every node stores sum of all nodes in left subtree
- Number of leaf nodes in the subtree of every node of an n-ary tree
- Implementing a BST where every node stores the maximum number of nodes in the path till any leaf
- Print the nodes of binary tree as they become the leaf node
- Distance between two nodes of binary tree with node values from 1 to N
- Check if two nodes are in same subtree of the root node
- Print all the nodes except the leftmost node in every level of the given binary tree
- Minimum and maximum node that lies in the path connecting two nodes in a Binary Tree
- Subtree of all nodes in a tree using DFS
- Find the Kth node in the DFS traversal of a given subtree in a Tree
- Subtree with given sum in a Binary Tree
- Check if a binary tree is subtree of another binary tree | Set 2
- Check if a binary tree is subtree of another binary tree | Set 1
- Find the largest BST subtree in a given Binary Tree | Set 1
- Check if the given Binary Tree have a Subtree with equal no of 1's and 0's
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
