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; } |
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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 |
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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) |
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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 |
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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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