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# Convert a given tree to its Sum Tree

• Difficulty Level : Medium
• Last Updated : 08 Sep, 2021

Given a Binary Tree where each node has positive and negative values. Convert this to a tree where each node contains the sum of the left and right sub trees in the original tree. The values of leaf nodes are changed to 0.

For example, the following tree

```                  10
/      \
-2        6
/   \      /  \
8     -4    7    5```

should be changed to

```                 20(4-2+12+6)
/      \
4(8-4)      12(7+5)
/   \      /  \
0      0    0    0```

Solution:
Do a traversal of the given tree. In the traversal, store the old value of the current node, recursively call for left and right subtrees and change the value of current node as sum of the values returned by the recursive calls. Finally return the sum of new value and value (which is sum of values in the subtree rooted with this node).

## C++

 `// C++ program to convert a tree into its sum tree``#include ``using` `namespace` `std;` `/* A tree node structure */``class` `node``{``    ``public``:``int` `data;``node *left;``node *right;``};` `// Convert a given tree to a tree where``// every node contains sum of values of``// nodes in left and right subtrees in the original tree``int` `toSumTree(node *Node)``{``    ``// Base case``    ``if``(Node == NULL)``    ``return` `0;` `    ``// Store the old value``    ``int` `old_val = Node->data;` `    ``// Recursively call for left and``    ``// right subtrees and store the sum as``    ``// old value of this node``    ``Node->data = toSumTree(Node->left) + toSumTree(Node->right);` `    ``// Return the sum of values of nodes``    ``// in left and right subtrees and``    ``// old_value of this node``    ``return` `Node->data + old_val;``}` `// A utility function to print``// inorder traversal of a Binary Tree``void` `printInorder(node* Node)``{``    ``if` `(Node == NULL)``        ``return``;``    ``printInorder(Node->left);``    ``cout<<``" "``<data;``    ``printInorder(Node->right);``}` `/* Utility function to create a new Binary Tree node */``node* newNode(``int` `data)``{``    ``node *temp = ``new` `node;``    ``temp->data = data;``    ``temp->left = NULL;``    ``temp->right = NULL;``    ` `    ``return` `temp;``}` `/* Driver code */``int` `main()``{``    ``node *root = NULL;``    ``int` `x;``    ` `    ``/* Constructing tree given in the above figure */``    ``root = newNode(10);``    ``root->left = newNode(-2);``    ``root->right = newNode(6);``    ``root->left->left = newNode(8);``    ``root->left->right = newNode(-4);``    ``root->right->left = newNode(7);``    ``root->right->right = newNode(5);``    ` `    ``toSumTree(root);``    ` `    ``// Print inorder traversal of the converted``    ``// tree to test result of toSumTree()``    ``cout<<``"Inorder Traversal of the resultant tree is: \n"``;``    ``printInorder(root);``    ``return` `0;``}` `// This code is contributed by rathbhupendra`

## C

 `#include` `/* A tree node structure */``struct` `node``{``  ``int` `data;``  ``struct` `node *left;``  ``struct` `node *right;``};` `// Convert a given tree to a tree where every node contains sum of values of``// nodes in left and right subtrees in the original tree``int` `toSumTree(``struct` `node *node)``{``    ``// Base case``    ``if``(node == NULL)``      ``return` `0;` `    ``// Store the old value``    ``int` `old_val = node->data;` `    ``// Recursively call for left and right subtrees and store the sum as``    ``// new value of this node``    ``node->data = toSumTree(node->left) + toSumTree(node->right);` `    ``// Return the sum of values of nodes in left and right subtrees and``    ``// old_value of this node``    ``return` `node->data + old_val;``}` `// A utility function to print inorder traversal of a Binary Tree``void` `printInorder(``struct` `node* node)``{``     ``if` `(node == NULL)``          ``return``;``     ``printInorder(node->left);``     ``printf``(``"%d "``, node->data);``     ``printInorder(node->right);``}` `/* Utility function to create a new Binary Tree node */``struct` `node* newNode(``int` `data)``{``  ``struct` `node *temp = ``new` `struct` `node;``  ``temp->data = data;``  ``temp->left = NULL;``  ``temp->right = NULL;` `  ``return` `temp;``}` `/* Driver function to test above functions */``int` `main()``{``  ``struct` `node *root = NULL;``  ``int` `x;` `  ``/* Constructing tree given in the above figure */``  ``root = newNode(10);``  ``root->left = newNode(-2);``  ``root->right = newNode(6);``  ``root->left->left = newNode(8);``  ``root->left->right = newNode(-4);``  ``root->right->left = newNode(7);``  ``root->right->right = newNode(5);` `  ``toSumTree(root);` `  ``// Print inorder traversal of the converted tree to test result of toSumTree()``  ``printf``(``"Inorder Traversal of the resultant tree is: \n"``);``  ``printInorder(root);` `  ``getchar``();``  ``return` `0;``}`

## Java

 `// Java program to convert a tree into its sum tree`` ` `// A binary tree node``class` `Node``{``    ``int` `data;``    ``Node left, right;`` ` `    ``Node(``int` `item)``    ``{``        ``data = item;``        ``left = right = ``null``;``    ``}``}`` ` `class` `BinaryTree``{``    ``Node root;`` ` `    ``// Convert a given tree to a tree where every node contains sum of``    ``// values of nodes in left and right subtrees in the original tree``    ``int` `toSumTree(Node node)``    ``{``        ``// Base case``        ``if` `(node == ``null``)``            ``return` `0``;`` ` `        ``// Store the old value``        ``int` `old_val = node.data;`` ` `        ``// Recursively call for left and right subtrees and store the sum``        ``// as new value of this node``        ``node.data = toSumTree(node.left) + toSumTree(node.right);`` ` `        ``// Return the sum of values of nodes in left and right subtrees``        ``// and old_value of this node``        ``return` `node.data + old_val;``    ``}`` ` `    ``// A utility function to print inorder traversal of a Binary Tree``    ``void` `printInorder(Node node)``    ``{``        ``if` `(node == ``null``)``            ``return``;``        ``printInorder(node.left);``        ``System.out.print(node.data + ``" "``);``        ``printInorder(node.right);``    ``}`` ` `    ``/* Driver function to test above functions */``    ``public` `static` `void` `main(String args[])``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();`` ` `        ``/* Constructing tree given in the above figure */``        ``tree.root = ``new` `Node(``10``);``        ``tree.root.left = ``new` `Node(-``2``);``        ``tree.root.right = ``new` `Node(``6``);``        ``tree.root.left.left = ``new` `Node(``8``);``        ``tree.root.left.right = ``new` `Node(-``4``);``        ``tree.root.right.left = ``new` `Node(``7``);``        ``tree.root.right.right = ``new` `Node(``5``);`` ` `        ``tree.toSumTree(tree.root);`` ` `        ``// Print inorder traversal of the converted tree to test result``        ``// of toSumTree()``        ``System.out.println(``"Inorder Traversal of the resultant tree is:"``);``        ``tree.printInorder(tree.root);``    ``}``}` `// This code has been contributed by Mayank Jaiswal`

## Python3

 `# Python3 program to convert a tree``# into its sum tree` `# Node definition``class` `node:``    ` `    ``def` `__init__(``self``, data):``        ``self``.left ``=` `None``        ``self``.right ``=` `None``        ``self``.data ``=` `data` `# Convert a given tree to a tree where``# every node contains sum of values of``# nodes in left and right subtrees``# in the original tree``def` `toSumTree(Node) :``    ` `    ``# Base case``    ``if``(Node ``=``=` `None``) :``        ``return` `0` `    ``# Store the old value``    ``old_val ``=` `Node.data` `    ``# Recursively call for left and``    ``# right subtrees and store the sum as``    ``# new value of this node``    ``Node.data ``=` `toSumTree(Node.left) ``+` `\``                ``toSumTree(Node.right)` `    ``# Return the sum of values of nodes``    ``# in left and right subtrees and``    ``# old_value of this node``    ``return` `Node.data ``+` `old_val` `# A utility function to print``# inorder traversal of a Binary Tree``def` `printInorder(Node) :``    ``if` `(Node ``=``=` `None``) :``        ``return``    ``printInorder(Node.left)``    ``print``(Node.data, end ``=` `" "``)``    ``printInorder(Node.right)``    ` `# Utility function to create a new Binary Tree node``def` `newNode(data) :``    ``temp ``=` `node(``0``)``    ``temp.data ``=` `data``    ``temp.left ``=` `None``    ``temp.right ``=` `None``    ` `    ``return` `temp` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``root ``=` `None``    ``x ``=` `0``    ` `    ``# Constructing tree given in the above figure``    ``root ``=` `newNode(``10``)``    ``root.left ``=` `newNode(``-``2``)``    ``root.right ``=` `newNode(``6``)``    ``root.left.left ``=` `newNode(``8``)``    ``root.left.right ``=` `newNode(``-``4``)``    ``root.right.left ``=` `newNode(``7``)``    ``root.right.right ``=` `newNode(``5``)``    ` `    ``toSumTree(root)``    ` `    ``# Print inorder traversal of the converted``    ``# tree to test result of toSumTree()``    ``print``(``"Inorder Traversal of the resultant tree is: "``)``    ``printInorder(root)` `# This code is contributed by Arnab Kundu`

## C#

 `// C# program to convert a tree``// into its sum tree``using` `System;` `// A binary tree node``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node left, right;` `    ``public` `Node(``int` `item)``    ``{``        ``data = item;``        ``left = right = ``null``;``    ``}``}` `class` `GFG``{``public` `Node root;` `// Convert a given tree to a tree where``// every node contains sum of values of``// nodes in left and right subtrees in``// the original tree``public` `virtual` `int` `toSumTree(Node node)``{``    ``// Base case``    ``if` `(node == ``null``)``    ``{``        ``return` `0;``    ``}` `    ``// Store the old value``    ``int` `old_val = node.data;` `    ``// Recursively call for left and``    ``// right subtrees and store the sum``    ``// as new value of this node``    ``node.data = toSumTree(node.left) +``                ``toSumTree(node.right);` `    ``// Return the sum of values of nodes``    ``// in left and right subtrees old_value``    ``// of this node``    ``return` `node.data + old_val;``}` `// A utility function to print``// inorder traversal of a Binary Tree``public` `virtual` `void` `printInorder(Node node)``{``    ``if` `(node == ``null``)``    ``{``        ``return``;``    ``}``    ``printInorder(node.left);``    ``Console.Write(node.data + ``" "``);``    ``printInorder(node.right);``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``GFG tree = ``new` `GFG();` `    ``/* Constructing tree given in``       ``the above figure */``    ``tree.root = ``new` `Node(10);``    ``tree.root.left = ``new` `Node(-2);``    ``tree.root.right = ``new` `Node(6);``    ``tree.root.left.left = ``new` `Node(8);``    ``tree.root.left.right = ``new` `Node(-4);``    ``tree.root.right.left = ``new` `Node(7);``    ``tree.root.right.right = ``new` `Node(5);` `    ``tree.toSumTree(tree.root);` `    ``// Print inorder traversal of the``    ``// converted tree to test result of toSumTree()``    ``Console.WriteLine(``"Inorder Traversal of "` `+``                     ``"the resultant tree is:"``);``    ``tree.printInorder(tree.root);``}``}` `// This code is contributed by Shrikant13`

## Javascript

 ``

Output:

```Inorder Traversal of the resultant tree is:
0 4 0 20 0 12 0```

Time Complexity: The solution involves a simple traversal of the given tree. So the time complexity is O(n) where n is the number of nodes in the given Binary Tree.