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# Print all the paths from root, with a specified sum in Binary tree

Given a Binary tree and a sum S, print all the paths, starting from root, that sums upto the given sum.

Note that this problem is different from root to leaf paths. Here path doesn’t need to end on a leaf node.

Examples:

```Input :
Input : sum = 8,
Root of tree
1
/   \
20      3
/    \
4       15
/  \     /  \
6    7   8    9
Output :
Path: 1 3 4

Input : sum = 38,
Root of tree
10
/     \
28       13
/     \
14       15
/   \     /  \
21   22   23   24
Output : Path found: 10 28
Path found: 10 13 15```

For this problem, preorder traversal is best suited as we have to add up a key value as we land on that node.
We start from the root and start traversing by preorder traversal, adding key value to the sum_so_far and checking whether it is equal to the required sum.
Also, as tree node doesn’t have a pointer pointing to its parent, we have to explicitly save from where we have moved. We use a vector path to store the path for this.
Every node in this path contributes to the sum_so_far.

Implementation:

## C++

 `// C++ program to print all paths beginning with``// root and sum as given sum``#include``using` `namespace` `std;` `// A Tree node``struct` `Node``{``    ``int` `key;``    ``struct` `Node *left, *right;``};` `// Utility function to create a new node``Node* newNode(``int` `key)``{``    ``Node* temp = ``new` `Node;``    ``temp->key = key;``    ``temp->left = temp->right = NULL;``    ``return` `(temp);``}`  `void` `printPathsUtil(Node* curr_node, ``int` `sum,``            ``int` `sum_so_far, vector<``int``> &path)``{``    ``if` `(curr_node == NULL)``        ``return``;` `    ``// add the current node's value``    ``sum_so_far += curr_node->key;` `    ``// add the value to path``    ``path.push_back(curr_node->key);` `    ``// print the required path``    ``if` `(sum_so_far == sum )``    ``{``        ``cout << ``"Path found: "``;``        ``for` `(``int` `i=0; ileft != NULL)``        ``printPathsUtil(curr_node->left, sum, sum_so_far, path);` `    ``// if right child exists``    ``if` `(curr_node->right != NULL)``        ``printPathsUtil(curr_node->right, sum, sum_so_far, path);`  `    ``// Remove last element from path``    ``// and move back to parent``    ``path.pop_back();``}` `// Wrapper over printPathsUtil``void` `printPaths(Node *root, ``int` `sum)``{``    ``vector<``int``> path;``    ``printPathsUtil(root, sum, 0, path);``}` `// Driver program``int` `main ()``{``    ``/* 10``    ``/     \``    ``28     13``        ``/     \``        ``14     15``        ``/ \     / \``    ``21 22 23 24*/``    ``Node *root = newNode(10);``    ``root->left = newNode(28);``    ``root->right = newNode(13);` `    ``root->right->left = newNode(14);``    ``root->right->right = newNode(15);` `    ``root->right->left->left = newNode(21);``    ``root->right->left->right = newNode(22);``    ``root->right->right->left = newNode(23);``    ``root->right->right->right = newNode(24);` `    ``int` `sum = 38;` `    ``printPaths(root, sum);` `    ``return` `0;``}`

## Java

 `// Java program to print all paths beginning``// with root and sum as given sum``import` `java.util.ArrayList;` `class` `Graph{` `// A Tree node``static` `class` `Node``{``    ``int` `key;``    ``Node left, right;``};` `// Utility function to create a new node``static` `Node newNode(``int` `key)``{``    ``Node temp = ``new` `Node();``    ``temp.key = key;``    ``temp.left = temp.right = ``null``;``    ``return` `(temp);``}` `static` `void` `printPathsUtil(Node curr_node, ``int` `sum,``                           ``int` `sum_so_far,``                           ``ArrayList path)``{``    ``if` `(curr_node == ``null``)``        ``return``;``        ` `    ``// Add the current node's value``    ``sum_so_far += curr_node.key;` `    ``// Add the value to path``    ``path.add(curr_node.key);` `    ``// Print the required path``    ``if` `(sum_so_far == sum)``    ``{``        ``System.out.print(``"Path found: "``);``        ``for``(``int` `i = ``0``; i < path.size(); i++)``            ``System.out.print(path.get(i) + ``" "``);``            ` `        ``System.out.println();``    ``}` `    ``// If left child exists``    ``if` `(curr_node.left != ``null``)``        ``printPathsUtil(curr_node.left, sum,``                       ``sum_so_far, path);` `    ``// If right child exists``    ``if` `(curr_node.right != ``null``)``        ``printPathsUtil(curr_node.right, sum,``                       ``sum_so_far, path);` `    ``// Remove last element from path``    ``// and move back to parent``    ``path.remove(path.size() - ``1``);``}` `// Wrapper over printPathsUtil``static` `void` `printPaths(Node root, ``int` `sum)``{``    ``ArrayList path = ``new` `ArrayList<>();``    ``printPathsUtil(root, sum, ``0``, path);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ` `    ``/*    10``       ``/     \``     ``28       13``           ``/     \``         ``14       15``        ``/   \     /  \``       ``21   22   23   24*/``    ``Node root = newNode(``10``);``    ``root.left = newNode(``28``);``    ``root.right = newNode(``13``);` `    ``root.right.left = newNode(``14``);``    ``root.right.right = newNode(``15``);` `    ``root.right.left.left = newNode(``21``);``    ``root.right.left.right = newNode(``22``);``    ``root.right.right.left = newNode(``23``);``    ``root.right.right.right = newNode(``24``);` `    ``int` `sum = ``38``;` `    ``printPaths(root, sum);``}``}` `// This code is contributed by sanjeev2552`

## Python3

 `# Python3 program to Print all the``# paths from root, with a specified``# sum in Binary tree``    ` `# Binary Tree Node``""" utility that allocates a newNode``with the given key """``class` `newNode:` `    ``# Construct to create a newNode``    ``def` `__init__(``self``, key):``        ``self``.key ``=` `key``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# This function prints all paths``# that have sum k``def` `printPathsUtil(curr_node, ``sum``,``                ``sum_so_far, path):` `    ``# empty node``    ``if` `(``not` `curr_node) :``        ``return``    ``sum_so_far ``+``=` `curr_node.key``    ` `    ``# add current node to the path``    ``path.append(curr_node.key)` `    ``# print the required path``    ``if` `(sum_so_far ``=``=` `sum` `) :``    ` `        ``print``(``"Path found:"``, end ``=` `" "``)``        ``for` `i ``in` `range``(``len``(path)):``            ``print``(path[i], end ``=` `" "``)` `        ``print``()``    ` `    ``# if left child exists``    ``if` `(curr_node.left !``=` `None``) :``        ``printPathsUtil(curr_node.left, ``sum``,``                    ``sum_so_far, path)` `    ``# if right child exists``    ``if` `(curr_node.right !``=` `None``) :``        ``printPathsUtil(curr_node.right, ``sum``,``                    ``sum_so_far, path)` `    ``# Remove the current element``    ``# from the path``    ``path.pop(``-``1``)` `# A wrapper over printKPathUtil()``def` `printPaths(root, ``sum``):` `    ``path ``=` `[]``    ``printPathsUtil(root, ``sum``, ``0``, path)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``""" 10``    ``/     \``    ``28     13``        ``/     \``        ``14     15``        ``/ \     / \``    ``21 22 23 24"""``    ``root ``=` `newNode(``10``)``    ``root.left ``=` `newNode(``28``)``    ``root.right ``=` `newNode(``13``)` `    ``root.right.left ``=` `newNode(``14``)``    ``root.right.right ``=` `newNode(``15``)` `    ``root.right.left.left ``=` `newNode(``21``)``    ``root.right.left.right ``=` `newNode(``22``)``    ``root.right.right.left ``=` `newNode(``23``)``    ``root.right.right.right ``=` `newNode(``24``)` `    ``sum` `=` `38` `    ``printPaths(root, ``sum``)` `# This code is contributed by``# Shubham Singh(SHUBHAMSINGH10)`

## C#

 `using` `System;``using` `System.Collections.Generic;` `class` `Graph``{``    ``// A Tree node``    ``public` `class` `Node``    ``{``        ``public` `int` `key;``        ``public` `Node left, right;``    ``}` `    ``// Utility function to create a new node``    ``public` `static` `Node newNode(``int` `key)``    ``{``        ``Node temp = ``new` `Node();``        ``temp.key = key;``        ``temp.left = temp.right = ``null``;``        ``return` `(temp);``    ``}` `    ``public` `static` `void` `printPathsUtil(Node curr_node, ``int` `sum,``                                      ``int` `sum_so_far,``                                      ``List<``int``> path)``    ``{``        ``if` `(curr_node == ``null``)``            ``return``;` `        ``// Add the current node's value``        ``sum_so_far += curr_node.key;` `        ``// Add the value to path``        ``path.Add(curr_node.key);` `        ``// Print the required path``        ``if` `(sum_so_far == sum)``        ``{``            ``Console.Write(``"Path found: "``);``            ``for` `(``int` `i = 0; i < path.Count; i++)``                ``Console.Write(path[i] + ``" "``);` `            ``Console.WriteLine();``        ``}` `        ``// If left child exists``        ``if` `(curr_node.left != ``null``)``            ``printPathsUtil(curr_node.left, sum,``                           ``sum_so_far, path);` `        ``// If right child exists``        ``if` `(curr_node.right != ``null``)``            ``printPathsUtil(curr_node.right, sum,``                           ``sum_so_far, path);` `        ``// Remove last element from path``        ``// and move back to parent``        ``path.RemoveAt(path.Count - 1);``    ``}` `    ``// Wrapper over printPathsUtil``    ``public` `static` `void` `printPaths(Node root, ``int` `sum)``    ``{``        ``List<``int``> path = ``new` `List<``int``>();``        ``printPathsUtil(root, sum, 0, path);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(``string``[] args)``    ``{` `        ``/*    10``           ``/     \``         ``28       13``               ``/     \``             ``14       15``            ``/   \     /  \``           ``21   22   23   24*/``        ``Node root = newNode(10);``        ``root.left = newNode(28);``        ``root.right = newNode(13);` `        ``root.right.left = newNode(14);``        ``root.right.right = newNode(15);` `        ``root.right.left.left = newNode(21);``        ``root.right.left.right = newNode(22);``        ``root.right.right.left = newNode(23);``        ``root.right.right.right = newNode(24);` `        ``int` `sum = 38;` `        ``printPaths(root, sum);``    ``}``}` `// This code is contributed by Potta Lokesh`

## Javascript

 ``

Output

```Path found: 10 28
Path found: 10 13 15 ```

Time Complexity: O(N^2), in the worst case, where N is the number of nodes in the tree. This is because we potentially traverse all nodes in the tree, and for each leaf node, we check the sum of the path, which takes O(N) time in the worst case.
Auxiliary Space: O(h), where h is the height of the binary tree. This is because the maximum amount of space used by the algorithm at any given time is the size of the path vector, which is at most equal to the height of the binary tree. This is because the path vector is only added to when we are traversing down the tree, and its size is reduced back to 0 when we are traversing back up the tree. Therefore, the space used by the algorithm is proportional to the height of the tree.

This article is contributed by Shubham Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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