# Convert an arbitrary Binary Tree to a tree that holds Children Sum Property

Question: Given an arbitrary binary tree, convert it to a binary tree that holds Children Sum Property. You can only increment data values in any node (You cannot change the structure of the tree and cannot decrement the value of any node).

For example, the below tree doesn’t hold the children sum property, convert it to a tree that holds the property.

```             50
/
/
7             2
/              /
/               /
3        5      1      30
```

Algorithm:
Traverse the given tree in post order to convert it, i.e., first change left and right children to hold the children sum property then change the parent node.

Let difference between node’s data and children sum be diff.

```     diff = node’s children sum - node’s data
```

If diff is 0 then nothing needs to be done.

If diff > 0 ( node’s data is smaller than node’s children sum) increment the node’s data by diff.

If diff < 0 (node’s data is greater than the node's children sum) then increment one child’s data. We can choose to increment either left or right child if they both are not NULL. Let us always first increment the left child. Incrementing a child changes the subtree’s children sum property so we need to change left subtree also. So we recursively increment the left child. If left child is empty then we recursively call increment() for right child.

Let us run the algorithm for the given example.

First convert the left subtree (increment 7 to 8).

```             50
/
/
8             2
/              /
/               /
3        5      1      30
```

Then convert the right subtree (increment 2 to 31)

```          50
/
/
8            31
/            /
/            /
3       5    1       30
```

Now convert the root, we have to increment left subtree for converting the root.

```          50
/
/
19           31
/            /
/            /
14      5     1       30
```

Please note the last step – we have incremented 8 to 19, and to fix the subtree we have incremented 3 to 14.

Implementation:

## C++

 `/* C++ Program to convert an aribitary ` `binary tree to a tree that hold ` `children sum property */` `#include ` `#include  ` `using` `namespace` `std; ` ` `  `class` `node  ` `{  ` `    ``public``: ` `    ``int` `data;  ` `    ``node* left;  ` `    ``node* right;  ` `     `  `    ``/* Constructor that allocates a new node  ` `    ``with the given data and NULL left and right  ` `    ``pointers. */` `    ``node(``int` `data) ` `    ``{ ` `        ``this``->data = data;  ` `        ``this``->left = NULL;  ` `        ``this``->right = NULL; ` `    ``} ` `};  ` ` `  `/* This function is used  ` `to increment left subtree */` `void` `increment(node* node, ``int` `diff);  ` ` `  `/* This function changes a tree  ` `to to hold children sum property */` `void` `convertTree(node* node)  ` `{  ` `    ``int` `left_data = 0, right_data = 0, diff;  ` `     `  `    ``/* If tree is empty or it's a leaf   ` `        ``node then return true */` `    ``if` `(node == NULL || (node->left == NULL && ` `                        ``node->right == NULL))  ` `        ``return``;  ` `    ``else` `    ``{  ` `        ``/* convert left and right subtrees */` `        ``convertTree(node->left);  ` `        ``convertTree(node->right);  ` `     `  `        ``/* If left child is not present then 0 is used  ` `        ``as data of left child */` `        ``if` `(node->left != NULL)  ` `        ``left_data = node->left->data;  ` `     `  `        ``/* If right child is not present then 0 is used  ` `        ``as data of right child */` `        ``if` `(node->right != NULL)  ` `        ``right_data = node->right->data;  ` `     `  `        ``/* get the diff of node's data and children sum */` `        ``diff = left_data + right_data - node->data;  ` `     `  `        ``/* If node's children sum is  ` `        ``greater than the node's data */` `        ``if` `(diff > 0)  ` `        ``node->data = node->data + diff;  ` `     `  `        ``/* THIS IS TRICKY --> If node's data ` `        ``is greater than children sum,  ` `        ``then increment subtree by diff */` `        ``if` `(diff < 0)  ` `        ``increment(node, -diff); ``// -diff is used to make diff positive  ` `    ``}  ` `}  ` ` `  `/* This function is used  ` `to increment subtree by diff */` `void` `increment(node* node, ``int` `diff)  ` `{  ` `    ``/* IF left child is not  ` `    ``NULL then increment it */` `    ``if``(node->left != NULL)  ` `    ``{  ` `        ``node->left->data = node->left->data + diff;  ` `     `  `        ``// Recursively call to fix  ` `        ``// the descendants of node->left  ` `        ``increment(node->left, diff);  ` `    ``}  ` `    ``else` `if` `(node->right != NULL) ``// Else increment right child  ` `    ``{  ` `        ``node->right->data = node->right->data + diff;  ` `     `  `        ``// Recursively call to fix  ` `        ``// the descendants of node->right  ` `        ``increment(node->right, diff);  ` `    ``}  ` `}  ` ` `  `/* Given a binary tree,  ` `printInorder() prints out its  ` `inorder traversal*/` `void` `printInorder(node* node)  ` `{  ` `    ``if` `(node == NULL)  ` `        ``return``;  ` `     `  `    ``/* first recur on left child */` `    ``printInorder(node->left);  ` `     `  `    ``/* then print the data of node */` `    ``cout<data<<``" "``; ` `     `  `    ``/* now recur on right child */` `    ``printInorder(node->right);  ` `}  ` ` `  `/* Driver code */` `int` `main()  ` `{  ` `    ``node *root = ``new` `node(50);  ` `    ``root->left = ``new` `node(7);  ` `    ``root->right = ``new` `node(2);  ` `    ``root->left->left = ``new` `node(3);  ` `    ``root->left->right = ``new` `node(5);  ` `    ``root->right->left = ``new` `node(1);  ` `    ``root->right->right = ``new` `node(30);  ` `     `  `    ``cout << ````" Inorder traversal before conversion: "``` `<< endl;  ` `    ``printInorder(root); ` `     `  `    ``convertTree(root);  ` `     `  `    ``cout << ````" Inorder traversal after conversion: "``` `<< endl;  ` `    ``printInorder(root);  ` `    ``return` `0;  ` `}  ` ` `  `// This code is contributed by rathbhupendra `

## C

 `/* Program to convert an aribitary binary tree to ` `   ``a tree that holds children sum property */` ` `  `#include ` `#include ` ` `  `struct` `node ` `{ ` `  ``int` `data; ` `  ``struct` `node* left; ` `  ``struct` `node* right; ` `}; ` ` `  `/* This function is used to increment left subtree */` `void` `increment(``struct` `node* node, ``int` `diff); ` ` `  `/* Helper function that allocates a new node ` ` ``with the given data and NULL left and right ` ` ``pointers. */` `struct` `node* newNode(``int` `data); ` ` `  `/* This function changes a tree to to hold children sum ` `   ``property */` `void` `convertTree(``struct` `node* node) ` `{ ` `  ``int` `left_data = 0,  right_data = 0, diff; ` ` `  `  ``/* If tree is empty or it's a leaf node then ` `     ``return true */` `  ``if` `(node == NULL || ` `     ``(node->left == NULL && node->right == NULL)) ` `    ``return``; ` `  ``else` `  ``{ ` `    ``/* convert left and right subtrees  */` `    ``convertTree(node->left); ` `    ``convertTree(node->right); ` ` `  `    ``/* If left child is not present then 0 is used ` `       ``as data of left child */` `    ``if` `(node->left != NULL) ` `      ``left_data = node->left->data; ` ` `  `    ``/* If right child is not present then 0 is used ` `      ``as data of right child */` `    ``if` `(node->right != NULL) ` `      ``right_data = node->right->data; ` ` `  `    ``/* get the diff of node's data and children sum */` `    ``diff = left_data + right_data - node->data; ` ` `  `    ``/* If node's children sum is greater than the node's data */` `    ``if` `(diff > 0) ` `       ``node->data = node->data + diff; ` ` `  `    ``/* THIS IS TRICKY --> If node's data is greater than children sum, ` `      ``then increment subtree by diff */` `    ``if` `(diff < 0) ` `      ``increment(node, -diff);  ``// -diff is used to make diff positive ` `  ``} ` `} ` ` `  `/* This function is used to increment subtree by diff */` `void` `increment(``struct` `node* node, ``int` `diff) ` `{ ` `  ``/* IF left child is not NULL then increment it */` `  ``if``(node->left != NULL) ` `  ``{ ` `    ``node->left->data = node->left->data + diff; ` ` `  `    ``// Recursively call to fix the descendants of node->left ` `    ``increment(node->left, diff);   ` `  ``} ` `  ``else` `if` `(node->right != NULL) ``// Else increment right child ` `  ``{ ` `    ``node->right->data = node->right->data + diff; ` ` `  `    ``// Recursively call to fix the descendants of node->right ` `    ``increment(node->right, diff); ` `  ``} ` `} ` ` `  `/* Given a binary tree, printInorder() prints out its ` `   ``inorder traversal*/` `void` `printInorder(``struct` `node* node) ` `{ ` `  ``if` `(node == NULL) ` `    ``return``; ` ` `  `  ``/* first recur on left child */` `  ``printInorder(node->left); ` ` `  `  ``/* then print the data of node */` `  ``printf``(``"%d "``, node->data); ` ` `  `  ``/* now recur on right child */` `  ``printInorder(node->right); ` `} ` ` `  `/* Helper function that allocates a new node ` ` ``with the given data and NULL left and right ` ` ``pointers. */` `struct` `node* newNode(``int` `data) ` `{ ` `  ``struct` `node* node = ` `      ``(``struct` `node*)``malloc``(``sizeof``(``struct` `node)); ` `  ``node->data = data; ` `  ``node->left = NULL; ` `  ``node->right = NULL; ` `  ``return``(node); ` `} ` ` `  `/* Driver program to test above functions */` `int` `main() ` `{ ` `  ``struct` `node *root = newNode(50); ` `  ``root->left        = newNode(7); ` `  ``root->right       = newNode(2); ` `  ``root->left->left  = newNode(3); ` `  ``root->left->right = newNode(5); ` `  ``root->right->left  = newNode(1); ` `  ``root->right->right = newNode(30); ` ` `  `  ``printf``(````" Inorder traversal before conversion "````); ` `  ``printInorder(root); ` ` `  `  ``convertTree(root); ` ` `  `  ``printf``(````" Inorder traversal after conversion "````); ` `  ``printInorder(root); ` ` `  `  ``getchar``(); ` `  ``return` `0; ` `} `

## Java

 `    ``// Java program to convert an arbitrary binary tree to a tree that holds ` `// children sum property ` `  `  `// A binary tree node ` `class` `Node  ` `{ ` `    ``int` `data; ` `    ``Node left, right; ` `  `  `    ``Node(``int` `item)  ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` `  `  `class` `BinaryTree  ` `{ ` `    ``Node root; ` `    ``/* This function changes a tree to to hold children sum ` `       ``property */` `  `  `    ``void` `convertTree(Node node)  ` `    ``{ ` `        ``int` `left_data = ``0``, right_data = ``0``, diff; ` `  `  `        ``/* If tree is empty or it's a leaf node then ` `         ``return true */` `        ``if` `(node == ``null` `                ``|| (node.left == ``null` `&& node.right == ``null``)) ` `            ``return``; ` `        ``else` `        ``{ ` `            ``/* convert left and right subtrees  */` `            ``convertTree(node.left); ` `            ``convertTree(node.right); ` `        `  `            ``/* If left child is not present then 0 is used ` `             ``as data of left child */` `            ``if` `(node.left != ``null``) ` `                ``left_data = node.left.data; ` `             `  `            ``/* If right child is not present then 0 is used ` `             ``as data of right child */` `            ``if` `(node.right != ``null``) ` `                ``right_data = node.right.data; ` `  `  `            ``/* get the diff of node's data and children sum */` `            ``diff = left_data + right_data - node.data; ` `  `  `            ``/* If node's children sum is greater than the node's data */` `            ``if` `(diff > ``0``) ` `                ``node.data = node.data + diff; ` `  `  `            ``/* THIS IS TRICKY --> If node's data is greater than children ` `               ``sum, then increment subtree by diff */` `            ``if` `(diff < ``0``) ` `             `  `                ``// -diff is used to make diff positive ` `                ``increment(node, -diff);   ` `        ``} ` `    ``} ` `  `  `    ``/* This function is used to increment subtree by diff */` `    ``void` `increment(Node node, ``int` `diff)  ` `    ``{ ` `        ``/* IF left child is not NULL then increment it */` `        ``if` `(node.left != ``null``)  ` `        ``{ ` `            ``node.left.data = node.left.data + diff; ` `  `  `            ``// Recursively call to fix the descendants of node->left ` `            ``increment(node.left, diff); ` `        ``}  ` `        ``else` `if` `(node.right != ``null``) ``// Else increment right child ` `        ``{ ` `            ``node.right.data = node.right.data + diff; ` `  `  `            ``// Recursively call to fix the descendants of node->right ` `            ``increment(node.right, diff); ` `        ``} ` `    ``} ` `  `  `    ``/* Given a binary tree, printInorder() prints out its ` `     ``inorder traversal*/` `    ``void` `printInorder(Node node)  ` `    ``{ ` `        ``if` `(node == ``null``) ` `            ``return``; ` `             `  `        ``/* first recur on left child */` `        ``printInorder(node.left); ` `  `  `        ``/* then print the data of node */` `        ``System.out.print(node.data + ``" "``); ` `  `  `        ``/* now recur on right child */` `        ``printInorder(node.right); ` `    ``} ` `  `  `    ``// Driver program to test above functions ` `    ``public` `static` `void` `main(String args[])  ` `    ``{ ` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``tree.root = ``new` `Node(``50``); ` `        ``tree.root.left = ``new` `Node(``7``); ` `        ``tree.root.right = ``new` `Node(``2``); ` `        ``tree.root.left.left = ``new` `Node(``3``); ` `        ``tree.root.left.right = ``new` `Node(``5``); ` `        ``tree.root.right.left = ``new` `Node(``1``); ` `        ``tree.root.right.right = ``new` `Node(``30``); ` `  `  `        ``System.out.println(``"Inorder traversal before conversion is :"``); ` `        ``tree.printInorder(tree.root); ` `  `  `        ``tree.convertTree(tree.root); ` `        ``System.out.println(``""``); ` `  `  `        ``System.out.println(``"Inorder traversal after conversion is :"``); ` `        ``tree.printInorder(tree.root); ` `  `  `    ``} ` `} ` ` `  `// This code has been contributed by Mayank Jaiswal(mayank_24) `

## Python3

 `# Program to convert an aribitary binary tree  ` `# to a tree that holds children sum property  ` ` `  `# Helper function that allocates a new  ` `# node with the given data and None  ` `# left and right poers.                                      ` `class` `newNode:  ` ` `  `    ``# Construct to create a new node  ` `    ``def` `__init__(``self``, key):  ` `        ``self``.data ``=` `key ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  `# This function changes a tree to  ` `# hold children sum property  ` `def` `convertTree(node): ` ` `  `    ``left_data ``=` `0` `    ``right_data ``=` `0` `    ``diff``=``0` ` `  `    ``# If tree is empty or it's a  ` `    ``# leaf node then return true  ` `    ``if` `(node ``=``=` `None` `or` `(node.left ``=``=` `None` `and` `                         ``node.right ``=``=` `None``)): ` `        ``return` `     `  `    ``else``: ` `         `  `        ``""" convert left and right subtrees """` `        ``convertTree(node.left) ` `        ``convertTree(node.right) ` ` `  `    ``""" If left child is not present then 0  ` `    ``is used as data of left child """` `    ``if` `(node.left !``=` `None``): ` `        ``left_data ``=` `node.left.data ` ` `  `    ``""" If right child is not present then 0  ` `    ``is used as data of right child """` `    ``if` `(node.right !``=` `None``): ` `        ``right_data ``=` `node.right.data ` ` `  `    ``""" get the diff of node's data  ` `        ``and children sum """` `    ``diff ``=` `left_data ``+` `right_data ``-` `node.data ` ` `  `    ``""" If node's children sum is greater  ` `        ``than the node's data """` `    ``if` `(diff > ``0``): ` `        ``node.data ``=` `node.data ``+` `diff ` ` `  `    ``""" THIS IS TRICKY -. If node's data is  ` `    ``greater than children sum, then increment ` `    ``subtree by diff """` `    ``if` `(diff < ``0``): ` `        ``increment(node, ``-``diff) ``# -diff is used to ` `                               ``# make diff positive ` ` `  `""" This function is used to increment ` `    ``subtree by diff """` `def` `increment(node, diff): ` ` `  `    ``""" IF left child is not None ` `        ``then increment it """` `    ``if``(node.left !``=` `None``): ` `        ``node.left.data ``=` `node.left.data ``+` `diff ` ` `  `        ``# Recursively call to fix the  ` `        ``# descendants of node.left ` `        ``increment(node.left, diff)  ` ` `  `    ``elif``(node.right !``=` `None``): ``# Else increment right child ` `        ``node.right.data ``=` `node.right.data ``+` `diff ` ` `  `        ``# Recursively call to fix the  ` `        ``# descendants of node.right ` `        ``increment(node.right, diff) ` ` `  `""" Given a binary tree, printInorder()  ` `prints out its inorder traversal"""` `def` `printInorder(node): ` ` `  `    ``if` `(node ``=``=` `None``): ` `        ``return` ` `  `    ``""" first recur on left child """` `    ``printInorder(node.left) ` ` `  `    ``""" then prthe data of node """` `    ``print``(node.data,end``=``" "``) ` ` `  `    ``""" now recur on right child """` `    ``printInorder(node.right) ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``root ``=` `newNode(``50``) ` `    ``root.left     ``=` `newNode(``7``) ` `    ``root.right     ``=` `newNode(``2``) ` `    ``root.left.left ``=` `newNode(``3``) ` `    ``root.left.right ``=` `newNode(``5``) ` `    ``root.right.left ``=` `newNode(``1``) ` `    ``root.right.right ``=` `newNode(``30``) ` ` `  `    ``print``(``"Inorder traversal before conversion"``) ` `    ``printInorder(root) ` ` `  `    ``convertTree(root) ` ` `  `    ``print``(````" Inorder traversal after conversion "````) ` `    ``printInorder(root) ` ` `  `# This code is contributed by ` `# Shubham Singh(SHUBHAMSINGH10) `

## C#

 `// C# program to convert an arbitrary  ` `// binary tree to a tree that holds  ` `// children sum property  ` `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; ` ` `  `/* This function changes a tree to  ` `hold children sum property */` `public` `virtual` `void` `convertTree(Node node) ` `{ ` `    ``int` `left_data = 0, right_data = 0, diff; ` ` `  `    ``/* If tree is empty or it's a leaf  ` `    ``node then return true */` `    ``if` `(node == ``null` `|| (node.left == ``null` `&&  ` `                         ``node.right == ``null``)) ` `    ``{ ` `        ``return``; ` `    ``} ` `    ``else` `    ``{ ` `        ``/* convert left and right subtrees */` `        ``convertTree(node.left); ` `        ``convertTree(node.right); ` ` `  `        ``/* If left child is not present then  ` `        ``0 is used as data of left child */` `        ``if` `(node.left != ``null``) ` `        ``{ ` `            ``left_data = node.left.data; ` `        ``} ` ` `  `        ``/* If right child is not present then  ` `        ``0 is used as data of right child */` `        ``if` `(node.right != ``null``) ` `        ``{ ` `            ``right_data = node.right.data; ` `        ``} ` ` `  `        ``/* get the diff of node's data  ` `           ``and children sum */` `        ``diff = left_data + right_data - node.data; ` ` `  `        ``/* If node's children sum is greater  ` `           ``than the node's data */` `        ``if` `(diff > 0) ` `        ``{ ` `            ``node.data = node.data + diff; ` `        ``} ` ` `  `        ``/* THIS IS TRICKY --> If node's data is  ` `        ``greater than children sum, then increment ` `        ``subtree by diff */` `        ``if` `(diff < 0) ` `        ``{ ` ` `  `            ``// -diff is used to make diff positive  ` `            ``increment(node, -diff); ` `        ``} ` `    ``} ` `} ` ` `  `/* This function is used to increment  ` `subtree by diff */` `public` `virtual` `void` `increment(Node node, ``int` `diff) ` `{ ` `    ``/* IF left child is not NULL then  ` `    ``increment it */` `    ``if` `(node.left != ``null``) ` `    ``{ ` `        ``node.left.data = node.left.data + diff; ` ` `  `        ``// Recursively call to fix the  ` `        ``// descendants of node->left  ` `        ``increment(node.left, diff); ` `    ``} ` `    ``else` `if` `(node.right != ``null``) ``// Else increment right child ` `    ``{ ` `        ``node.right.data = node.right.data + diff; ` ` `  `        ``// Recursively call to fix the  ` `        ``// descendants of node->right  ` `        ``increment(node.right, diff); ` `    ``} ` `} ` ` `  `/* Given a binary tree, printInorder()  ` `prints out its inorder traversal*/` `public` `virtual` `void` `printInorder(Node node) ` `{ ` `    ``if` `(node == ``null``) ` `    ``{ ` `        ``return``; ` `    ``} ` ` `  `    ``/* first recur on left child */` `    ``printInorder(node.left); ` ` `  `    ``/* then print the data of node */` `    ``Console.Write(node.data + ``" "``); ` ` `  `    ``/* now recur on right child */` `    ``printInorder(node.right); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``GFG tree = ``new` `GFG(); ` `    ``tree.root = ``new` `Node(50); ` `    ``tree.root.left = ``new` `Node(7); ` `    ``tree.root.right = ``new` `Node(2); ` `    ``tree.root.left.left = ``new` `Node(3); ` `    ``tree.root.left.right = ``new` `Node(5); ` `    ``tree.root.right.left = ``new` `Node(1); ` `    ``tree.root.right.right = ``new` `Node(30); ` ` `  `    ``Console.WriteLine(``"Inorder traversal "` `+  ` `                  ``"before conversion is :"``); ` `    ``tree.printInorder(tree.root); ` ` `  `    ``tree.convertTree(tree.root); ` `    ``Console.WriteLine(``""``); ` ` `  `    ``Console.WriteLine(``"Inorder traversal "` `+  ` `                   ``"after conversion is :"``); ` `    ``tree.printInorder(tree.root); ` `} ` `} ` ` `  `// This code is contributed by Shrikant13 `

Output :

```Inorder traversal before conversion is :
3 7 5 50 1 2 30
Inorder traversal after conversion is :
14 19 5 50 1 31 30
```

Time Complexity: O(n^2), Worst case complexity is for a skewed tree such that nodes are in decreasing order from root to leaf.

Please write comments if you find any bug in the above algorithm or a better way to solve the same problem.

Tree Tree