# Given a binary tree, print all root-to-leaf paths

• Difficulty Level : Easy
• Last Updated : 24 Aug, 2022
```For the below example tree, all root-to-leaf paths are:
10 â€“> 8 â€“> 3
10 â€“> 8 â€“> 5
10 â€“> 2 â€“> 2```

Recommended Practice

Algorithm:
Use a path array path[] to store current root to leaf path. Traverse from root to all leaves in top-down fashion. While traversing, store data of all nodes in current path in array path[]. When we reach a leaf node, print the path array.

## C++

 `#include ``using` `namespace` `std;` `/* A binary tree node has data, pointer to left child``and a pointer to right child */``class` `node``{``    ``public``:``    ``int` `data;``    ``node* left;``    ``node* right;``};` `/* Prototypes for functions needed in printPaths() */``void` `printPathsRecur(node* node, ``int` `path[], ``int` `pathLen);``void` `printArray(``int` `ints[], ``int` `len);` `/*Given a binary tree, print out all of its root-to-leaf``paths, one per line. Uses a recursive helper to do the work.*/``void` `printPaths(node* node)``{``    ``int` `path[1000];``    ``printPathsRecur(node, path, 0);``}` `/* Recursive helper function -- given a node,``and an array containing the path from the root``node up to but not including this node,``print out all the root-leaf paths.*/``void` `printPathsRecur(node* node, ``int` `path[], ``int` `pathLen)``{``    ``if` `(node == NULL)``        ``return``;``    ` `    ``/* append this node to the path array */``    ``path[pathLen] = node->data;``    ``pathLen++;``    ` `    ``/* it's a leaf, so print the path that lead to here */``    ``if` `(node->left == NULL && node->right == NULL)``    ``{``        ``printArray(path, pathLen);``    ``}``    ``else``    ``{``        ``/* otherwise try both subtrees */``        ``printPathsRecur(node->left, path, pathLen);``        ``printPathsRecur(node->right, path, pathLen);``    ``}``}`  `/* UTILITY FUNCTIONS */``/* Utility that prints out an array on a line. */``void` `printArray(``int` `ints[], ``int` `len)``{``    ``int` `i;``    ``for` `(i = 0; i < len; i++)``    ``{``        ``cout << ints[i] << ``" "``;``    ``}``    ``cout<data = data;``    ``Node->left = NULL;``    ``Node->right = NULL;``    ` `    ``return``(Node);``}` `/* Driver code*/``int` `main()``{``    ` `    ``/* Constructed binary tree is``                ``10``            ``/ \``            ``8 2``        ``/ \ /``        ``3 5 2``    ``*/``    ``node *root = newnode(10);``    ``root->left = newnode(8);``    ``root->right = newnode(2);``    ``root->left->left = newnode(3);``    ``root->left->right = newnode(5);``    ``root->right->left = newnode(2);``    ` `    ``printPaths(root);``    ``return` `0;``}` `// This code is contributed by rathbhupendra`

## C

 `#include``#include`` ` `/* A binary tree node has data, pointer to left child``   ``and a pointer to right child */``struct` `node``{``   ``int` `data;``   ``struct` `node* left;``   ``struct` `node* right;``};` `/* Prototypes for functions needed in printPaths() */``void` `printPathsRecur(``struct` `node* node, ``int` `path[], ``int` `pathLen);``void` `printArray(``int` `ints[], ``int` `len);` `/*Given a binary tree, print out all of its root-to-leaf`` ``paths, one per line. Uses a recursive helper to do the work.*/``void` `printPaths(``struct` `node* node)``{``  ``int` `path[1000];``  ``printPathsRecur(node, path, 0);``}` `/* Recursive helper function -- given a node, and an array containing`` ``the path from the root node up to but not including this node,`` ``print out all the root-leaf paths.*/``void` `printPathsRecur(``struct` `node* node, ``int` `path[], ``int` `pathLen)``{``  ``if` `(node==NULL)``    ``return``;` `  ``/* append this node to the path array */``  ``path[pathLen] = node->data;``  ``pathLen++;` `  ``/* it's a leaf, so print the path that lead to here  */``  ``if` `(node->left==NULL && node->right==NULL)``  ``{``    ``printArray(path, pathLen);``  ``}``  ``else``  ``{``    ``/* otherwise try both subtrees */``    ``printPathsRecur(node->left, path, pathLen);``    ``printPathsRecur(node->right, path, pathLen);``  ``}``}`  `/* UTILITY FUNCTIONS */``/* Utility that prints out an array on a line. */``void` `printArray(``int` `ints[], ``int` `len)``{``  ``int` `i;``  ``for` `(i=0; idata = data;``  ``node->left = NULL;``  ``node->right = NULL;`` ` `  ``return``(node);``}`` ` `/* Driver program to test above functions*/``int` `main()``{`` ` `  ``/* Constructed binary tree is``            ``10``          ``/   \``        ``8      2``      ``/  \    /``    ``3     5  2``  ``*/``  ``struct` `node *root = newnode(10);``  ``root->left        = newnode(8);``  ``root->right       = newnode(2);``  ``root->left->left  = newnode(3);``  ``root->left->right = newnode(5);``  ``root->right->left = newnode(2);`` ` `  ``printPaths(root);`` ` `  ``getchar``();``  ``return` `0;``}`

## Java

 `// Java program to print all the node to leaf path`` ` `/* A binary tree node has data, pointer to left child``   ``and a pointer to right child */``class` `Node``{``    ``int` `data;``    ``Node left, right;`` ` `    ``Node(``int` `item)``    ``{``        ``data = item;``        ``left = right = ``null``;``    ``}``}`` ` `class` `BinaryTree``{``    ``Node root;``     ` `    ``/*Given a binary tree, print out all of its root-to-leaf``      ``paths, one per line. Uses a recursive helper to do``      ``the work.*/``    ``void` `printPaths(Node node)``    ``{``        ``int` `path[] = ``new` `int``[``1000``];``        ``printPathsRecur(node, path, ``0``);``    ``}`` ` `    ``/* Recursive helper function -- given a node, and an array``       ``containing the path from the root node up to but not``       ``including this node, print out all the root-leaf paths.*/``    ``void` `printPathsRecur(Node node, ``int` `path[], ``int` `pathLen)``    ``{``        ``if` `(node == ``null``)``            ``return``;`` ` `        ``/* append this node to the path array */``        ``path[pathLen] = node.data;``        ``pathLen++;`` ` `        ``/* it's a leaf, so print the path that lead to here  */``        ``if` `(node.left == ``null` `&& node.right == ``null``)``            ``printArray(path, pathLen);``        ``else``        ``{``            ``/* otherwise try both subtrees */``            ``printPathsRecur(node.left, path, pathLen);``            ``printPathsRecur(node.right, path, pathLen);``        ``}``    ``}`` ` `    ``/* Utility function that prints out an array on a line. */``    ``void` `printArray(``int` `ints[], ``int` `len)``    ``{``        ``int` `i;``        ``for` `(i = ``0``; i < len; i++)``        ``{``            ``System.out.print(ints[i] + ``" "``);``        ``}``        ``System.out.println(``""``);``    ``}`` ` `    ``// driver program to test above functions``    ``public` `static` `void` `main(String args[])``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();``        ``tree.root = ``new` `Node(``10``);``        ``tree.root.left = ``new` `Node(``8``);``        ``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(``2``);``        ` `        ``/* Let us test the built tree by printing Inorder traversal */``        ``tree.printPaths(tree.root);``    ``}``}` `// This code has been contributed by Mayank Jaiswal`

## Python3

 `"""``Python program to print all path from root to``leaf in a binary tree``"""` `# binary tree node contains data field ,``# left and right pointer``class` `Node:``    ``# constructor to create tree node``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# function to print all path from root``# to leaf in binary tree``def` `printPaths(root):``    ``# list to store path``    ``path ``=` `[]``    ``printPathsRec(root, path, ``0``)` `# Helper function to print path from root``# to leaf in binary tree``def` `printPathsRec(root, path, pathLen):``    ` `    ``# Base condition - if binary tree is``    ``# empty return``    ``if` `root ``is` `None``:``        ``return` `    ``# add current root's data into``    ``# path_ar list``    ` `    ``# if length of list is gre``    ``if``(``len``(path) > pathLen):``        ``path[pathLen] ``=` `root.data``    ``else``:``        ``path.append(root.data)` `    ``# increment pathLen by 1``    ``pathLen ``=` `pathLen ``+` `1` `    ``if` `root.left ``is` `None` `and` `root.right ``is` `None``:``        ` `        ``# leaf node then print the list``        ``printArray(path, pathLen)``    ``else``:``        ``# try for left and right subtree``        ``printPathsRec(root.left, path, pathLen)``        ``printPathsRec(root.right, path, pathLen)` `# Helper function to print list in which``# root-to-leaf path is stored``def` `printArray(ints, ``len``):``    ``for` `i ``in` `ints[``0` `: ``len``]:``        ``print``(i,``" "``,end``=``"")``    ``print``()` `# Driver program to test above function``"""``Constructed binary tree is``            ``10``        ``/ \``        ``8     2``    ``/ \ /``    ``3 5 2``"""``root ``=` `Node(``10``)``root.left ``=` `Node(``8``)``root.right ``=` `Node(``2``)``root.left.left ``=` `Node(``3``)``root.left.right ``=` `Node(``5``)``root.right.left ``=` `Node(``2``)``printPaths(root)` `# This code has been contributed by Shweta Singh.`

## C#

 `using` `System;` `// C# program to print all the node to leaf path` `/* A binary tree node has data, pointer to left child``   ``and a pointer to right child */``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node left, right;` `    ``public` `Node(``int` `item)``    ``{``        ``data = item;``        ``left = right = ``null``;``    ``}``}` `public` `class` `BinaryTree``{``    ``public` `Node root;` `    ``/*Given a binary tree, print out all of its root-to-leaf``      ``paths, one per line. Uses a recursive helper to do ``      ``the work.*/``    ``public` `virtual` `void` `printPaths(Node node)``    ``{``        ``int``[] path = ``new` `int``[1000];``        ``printPathsRecur(node, path, 0);``    ``}` `    ``/* Recursive helper function -- given a node, and an array``       ``containing the path from the root node up to but not ``       ``including this node, print out all the root-leaf paths.*/``    ``public` `virtual` `void` `printPathsRecur(Node node, ``int``[] path, ``int` `pathLen)``    ``{``        ``if` `(node == ``null``)``        ``{``            ``return``;``        ``}` `        ``/* append this node to the path array */``        ``path[pathLen] = node.data;``        ``pathLen++;` `        ``/* it's a leaf, so print the path that lead to here  */``        ``if` `(node.left == ``null` `&& node.right == ``null``)``        ``{``            ``printArray(path, pathLen);``        ``}``        ``else``        ``{``            ``/* otherwise try both subtrees */``            ``printPathsRecur(node.left, path, pathLen);``            ``printPathsRecur(node.right, path, pathLen);``        ``}``    ``}` `    ``/* Utility function that prints out an array on a line. */``    ``public` `virtual` `void` `printArray(``int``[] ints, ``int` `len)``    ``{``        ``int` `i;``        ``for` `(i = 0; i < len; i++)``        ``{``            ``Console.Write(ints[i] + ``" "``);``        ``}``        ``Console.WriteLine(``""``);``    ``}` `    ``// driver program to test above functions``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();``        ``tree.root = ``new` `Node(10);``        ``tree.root.left = ``new` `Node(8);``        ``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(2);` `        ``/* Let us test the built tree by printing Inorder traversal */``        ``tree.printPaths(tree.root);``    ``}``}` `// This code is contributed by Shrikant13`

## Javascript

 ``

Output

```10 8 3
10 8 5
10 2 2 ```

Time Complexity: O(n) where n is number of nodes.

Space complexity: O(h) where h is height Of a Binary Tree.

Another Method

## C++

 `#include``using` `namespace` `std;``/*Binary Tree representation using structure where data is in integer and 2 pointer``struct Node* left and struct Node* right represents left and right pointers of a node``in a tree respectively*/``struct` `Node``{``    ``int` `data;``    ``struct` `Node* left;``    ``struct` `Node* right;` `    ``Node(``int` `x){``        ``data = x;``        ``left = right = NULL;``    ``}``};``/*Recursive helper function which will recursively update our ans vector.``it takes root of our tree ,arr vector and ans vector as an argument*/` `void` `helper(Node* root,vector<``int``> arr,vector> &ans)``{``    ``if``(!root)``        ``return``;``    ``arr.push_back(root->data);``    ``if``(root->left==NULL && root->right==NULL)``    ``{``       ``/*This will be only true when the node is leaf node``         ``and hence we will update our ans vector by inserting``         ``vector arr which have one unique path from root to leaf*/``       ``ans.push_back(arr);``       ``//after that we will return since we don't want to check after leaf node``        ``return``;``    ``}``    ``/*recursively going left and right until we find the leaf and updating the arr``    ``and ans vector simultaneously*/``    ``helper(root->left,arr,ans);``    ``helper(root->right,arr,ans);``}``vector> Paths(Node* root)``{``    ``/*creating 2-d vector in which each element is a 1-d vector``      ``having one unique path from root to leaf*/``    ``vector> ans;``    ``/*if root is null then there is no further action require so return*/``    ``if``(!root)``        ``return` `ans;``    ``vector<``int``> arr;``    ``/*arr is a vector which will have one unique path from root to leaf``     ``at a time.arr will be updated recursively*/``    ``helper(root,arr,ans);``    ``/*after helper function call our ans vector updated with paths so we will return ans vector*/``    ``return` `ans;``}``int`  `main()``{``   ``/*defining root and our tree*/``    ``Node *root = ``new` `Node(10);``    ``root->left = ``new` `Node(8);``    ``root->right = ``new` `Node(2);``    ``root->left->left = ``new` `Node(3);``    ``root->left->right = ``new` `Node(5);``    ``root->right->left = ``new` `Node(2);``   ``/*The answer returned will be stored in final 2-d vector*/``   ``vector> final=Paths(root);``   ``/*printing paths from root to leaf*/``   ``for``(``int` `i=0;i

## Python3

 `"""``Python program to print all path from root to``leaf in a binary tree``"""` `# binary tree node contains data field ,``# left and right pointer`  `class` `Node:``    ``# constructor to create tree node``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Recursive helper function which will recursively update our ans array.``# it takes root of our tree ,arr array and ans array as an argument`  `def` `helper(root, arr, ans):``    ``if` `not` `root:``        ``return` `    ``arr.append(root.data)` `    ``if` `root.left ``is` `None` `and` `root.right ``is` `None``:``        ``# This will be only true when the node is leaf node``        ``# and hence we will update our ans array by inserting``        ``# array arr which have one unique path from root to leaf``        ``ans.append(arr.copy())``        ``del` `arr[``-``1``]``        ``# after that we will return since we don't want to check after leaf node``        ``return` `    ``# recursively going left and right until we find the leaf and updating the arr``    ``# and ans array simultaneously``    ``helper(root.left, arr, ans)``    ``helper(root.right, arr, ans)``    ``del` `arr[``-``1``]`  `def` `Paths(root):``    ``# creating answer in which each element is a array``    ``# having one unique path from root to leaf``    ``ans ``=` `[]``    ``# if root is null then there is no further action require so return``    ``if` `not` `root:``        ``return` `[[]]``    ``arr ``=` `[]``    ``# arr is a array which will have one unique path from root to leaf``    ``# at a time.arr will be updated recursively``    ``helper(root, arr, ans)``    ``# after helper function call our ans array updated with paths so we will return ans array``    ``return` `ans`  `# Helper function to print list in which``# root-to-leaf path is stored``def` `printArray(paths):``    ``for` `path ``in` `paths:``        ``for` `i ``in` `path:``            ``print``(i, ``" "``, end``=``"")``        ``print``()`  `# Driver program to test above function``"""``Constructed binary tree is``            ``10``        ``/ \``        ``8     2``    ``/ \ /``    ``3 5 2``"""``root ``=` `Node(``10``)``root.left ``=` `Node(``8``)``root.right ``=` `Node(``2``)``root.left.left ``=` `Node(``3``)``root.left.right ``=` `Node(``5``)``root.right.left ``=` `Node(``2``)``paths ``=` `Paths(root)``printArray(paths)` `# This Code is Contributed by Vivek Maddeshiya`

Output

```10 8 3
10 8 5
10 2 2 ```

Time complexity: O(n)
Auxiliary Space : O(h) where h is the height of the binary tree.
Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem.

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