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# Print all k-sum paths in a binary tree

• Difficulty Level : Hard
• Last Updated : 13 Jul, 2021

A binary tree and a number k are given. Print every path in the tree with sum of the nodes in the path as k.
A path can start from any node and end at any node and must be downward only, i.e. they need not be root node and leaf node; and negative numbers can also be there in the tree.
Examples:

```Input : k = 5
Root of below binary tree:
1
/     \
3        -1
/   \     /   \
2     1   4     5
/   / \     \
1   1   2     6

Output :
3 2
3 1 1
1 3 1
4 1
1 -1 4 1
-1 4 2
5
1 -1 5 ```

Kindly note that this problem is significantly different from finding k-sum path from root to leaves. Here each node can be treated as root, hence the path can start and end at any node.
The basic idea to solve the problem is to do a preorder traversal of the given tree. We also need a container (vector) to keep track of the path that led to that node. At each node we check if there are any path that sums to k, if any we print the path and proceed recursively to print each path.
Below is the implementation of the same.

## C++

 `// C++ program to print all paths with sum k.``#include ``using` `namespace` `std;` `//utility function to print contents of``//a vector from index i to it's end``void` `printVector(``const` `vector<``int``>& v, ``int` `i)``{``    ``for` `(``int` `j=i; j& path,``                                           ``int` `k)``{``    ``// empty node``    ``if` `(!root)``        ``return``;` `    ``// add current node to the path``    ``path.push_back(root->data);` `    ``// check if there's any k sum path``    ``// in the left sub-tree.``    ``printKPathUtil(root->left, path, k);` `    ``// check if there's any k sum path``    ``// in the right sub-tree.``    ``printKPathUtil(root->right, path, k);` `    ``// check if there's any k sum path that``    ``// terminates at this node``    ``// Traverse the entire path as``    ``// there can be negative elements too``    ``int` `f = 0;``    ``for` `(``int` `j=path.size()-1; j>=0; j--)``    ``{``        ``f += path[j];` `        ``// If path sum is k, print the path``        ``if` `(f == k)``            ``printVector(path, j);``    ``}` `    ``// Remove the current element from the path``    ``path.pop_back();``}` `// A wrapper over printKPathUtil()``void` `printKPath(Node *root, ``int` `k)``{``    ``vector<``int``> path;``    ``printKPathUtil(root, path, k);``}` `// Driver code``int` `main()``{``    ``Node *root = ``new` `Node(1);``    ``root->left = ``new` `Node(3);``    ``root->left->left = ``new` `Node(2);``    ``root->left->right = ``new` `Node(1);``    ``root->left->right->left = ``new` `Node(1);``    ``root->right = ``new` `Node(-1);``    ``root->right->left = ``new` `Node(4);``    ``root->right->left->left = ``new` `Node(1);``    ``root->right->left->right = ``new` `Node(2);``    ``root->right->right = ``new` `Node(5);``    ``root->right->right->right = ``new` `Node(2);` `    ``int` `k = 5;``    ``printKPath(root, k);` `    ``return` `0;``}`

## Java

 `// Java program to print all paths with sum k.``import` `java.util.*;` `class` `GFG``{``    ` `//utility function to print contents of``//a vector from index i to it's end``static` `void` `printVector( Vector v, ``int` `i)``{``    ``for` `(``int` `j = i; j < v.size(); j++)``        ``System.out.print( v.get(j) + ``" "``);``        ``System.out.println();``}` `// binary tree node``static` `class` `Node``{``    ``int` `data;``    ``Node left,right;``    ``Node(``int` `x)``    ``{``        ``data = x;``        ``left = right = ``null``;``    ``}``};``static` `Vector path = ``new` `Vector();` `// This function prints all paths that have sum k``static` `void` `printKPathUtil(Node root, ``int` `k)``{``    ``// empty node``    ``if` `(root == ``null``)``        ``return``;` `    ``// add current node to the path``    ``path.add(root.data);` `    ``// check if there's any k sum path``    ``// in the left sub-tree.``    ``printKPathUtil(root.left, k);` `    ``// check if there's any k sum path``    ``// in the right sub-tree.``    ``printKPathUtil(root.right, k);` `    ``// check if there's any k sum path that``    ``// terminates at this node``    ``// Traverse the entire path as``    ``// there can be negative elements too``    ``int` `f = ``0``;``    ``for` `(``int` `j = path.size() - ``1``; j >= ``0``; j--)``    ``{``        ``f += path.get(j);` `        ``// If path sum is k, print the path``        ``if` `(f == k)``            ``printVector(path, j);``    ``}` `    ``// Remove the current element from the path``    ``path.remove(path.size() - ``1``);``}` `// A wrapper over printKPathUtil()``static` `void` `printKPath(Node root, ``int` `k)``{``    ``path = ``new` `Vector();``    ``printKPathUtil(root, k);``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``Node root = ``new` `Node(``1``);``    ``root.left = ``new` `Node(``3``);``    ``root.left.left = ``new` `Node(``2``);``    ``root.left.right = ``new` `Node(``1``);``    ``root.left.right.left = ``new` `Node(``1``);``    ``root.right = ``new` `Node(-``1``);``    ``root.right.left = ``new` `Node(``4``);``    ``root.right.left.left = ``new` `Node(``1``);``    ``root.right.left.right = ``new` `Node(``2``);``    ``root.right.right = ``new` `Node(``5``);``    ``root.right.right.right = ``new` `Node(``2``);` `    ``int` `k = ``5``;``    ``printKPath(root, k);``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 program to print all paths``# with sum k` `# utility function to print contents of``# a vector from index i to it's end``def` `printVector(v, i):``    ``for` `j ``in` `range``(i, ``len``(v)):``        ``print``(v[j], end ``=` `" "``)``    ``print``()``    ` `# Binary Tree Node``""" utility that allocates a newNode``with the given key """``class` `newNode:` `    ``# Construct to create a newNode``    ``def` `__init__(``self``, key):``        ``self``.data ``=` `key``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# This function prints all paths``# that have sum k``def` `printKPathUtil(root, path, k):` `    ``# empty node``    ``if` `(``not` `root) :``        ``return` `    ``# add current node to the path``    ``path.append(root.data)` `    ``# check if there's any k sum path``    ``# in the left sub-tree.``    ``printKPathUtil(root.left, path, k)` `    ``# check if there's any k sum path``    ``# in the right sub-tree.``    ``printKPathUtil(root.right, path, k)` `    ``# check if there's any k sum path that``    ``# terminates at this node``    ``# Traverse the entire path as``    ``# there can be negative elements too``    ``f ``=` `0``    ``for` `j ``in` `range``(``len``(path) ``-` `1``, ``-``1``, ``-``1``):    ``        ``f ``+``=` `path[j]` `        ``# If path sum is k, print the path``        ``if` `(f ``=``=` `k) :``            ``printVector(path, j)``    ` `    ``# Remove the current element``    ``# from the path``    ``path.pop(``-``1``)` `# A wrapper over printKPathUtil()``def` `printKPath(root, k):` `    ``path ``=``[]``    ``printKPathUtil(root, path, k)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``root ``=` `newNode(``1``)``    ``root.left ``=` `newNode(``3``)``    ``root.left.left ``=` `newNode(``2``)``    ``root.left.right ``=` `newNode(``1``)``    ``root.left.right.left ``=` `newNode(``1``)``    ``root.right ``=` `newNode(``-``1``)``    ``root.right.left ``=` `newNode(``4``)``    ``root.right.left.left ``=` `newNode(``1``)``    ``root.right.left.right ``=` `newNode(``2``)``    ``root.right.right ``=` `newNode(``5``)``    ``root.right.right.right ``=` `newNode(``2``)` `    ``k ``=` `5``    ``printKPath(root, k)` `# This code is contributed by``# Shubham Singh(SHUBHAMSINGH10)`

## C#

 `// C# program to print all paths with sum k.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `//utility function to print contents of``//a vector from index i to it's end``static` `void` `printList(List<``int``> v, ``int` `i)``{``    ``for` `(``int` `j = i; j < v.Count; j++)``        ``Console.Write(v[j] + ``" "``);``        ``Console.WriteLine();``}` `// binary tree node``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node left,right;``    ``public` `Node(``int` `x)``    ``{``        ``data = x;``        ``left = right = ``null``;``    ``}``};``static` `List<``int``> path = ``new` `List<``int``>();` `// This function prints all paths that have sum k``static` `void` `printKPathUtil(Node root, ``int` `k)``{``    ``// empty node``    ``if` `(root == ``null``)``        ``return``;` `    ``// add current node to the path``    ``path.Add(root.data);` `    ``// check if there's any k sum path``    ``// in the left sub-tree.``    ``printKPathUtil(root.left, k);` `    ``// check if there's any k sum path``    ``// in the right sub-tree.``    ``printKPathUtil(root.right, k);` `    ``// check if there's any k sum path that``    ``// terminates at this node``    ``// Traverse the entire path as``    ``// there can be negative elements too``    ``int` `f = 0;``    ``for` `(``int` `j = path.Count - 1; j >= 0; j--)``    ``{``        ``f += path[j];` `        ``// If path sum is k, print the path``        ``if` `(f == k)``            ``printList(path, j);``    ``}` `    ``// Remove the current element from the path``    ``path.RemoveAt(path.Count - 1);``}` `// A wrapper over printKPathUtil()``static` `void` `printKPath(Node root, ``int` `k)``{``    ``path = ``new` `List<``int``>();``    ``printKPathUtil(root, k);``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``Node root = ``new` `Node(1);``    ``root.left = ``new` `Node(3);``    ``root.left.left = ``new` `Node(2);``    ``root.left.right = ``new` `Node(1);``    ``root.left.right.left = ``new` `Node(1);``    ``root.right = ``new` `Node(-1);``    ``root.right.left = ``new` `Node(4);``    ``root.right.left.left = ``new` `Node(1);``    ``root.right.left.right = ``new` `Node(2);``    ``root.right.right = ``new` `Node(5);``    ``root.right.right.right = ``new` `Node(2);` `    ``int` `k = 5;``    ``printKPath(root, k);``}``}` `// This code is contributed by PrinciRaj1992`

Output:

```3 2
3 1 1
1 3 1
4 1
1 -1 4 1
-1 4 2
5
1 -1 5 ```

Time Complexity: O(n*h*h)  , as maximum size of path vector can be h

Space Complexity: O(h)

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