Given a binary tree, print out all of its root-to-leaf paths one per line.

Given the roots of a tree. print out all of its root-to-leaf paths one per line.. 
Algorithm:

initialize: pathlen = 0, path[1000] 
/*1000 is some max limit for paths, it can change*/

/*printPathsRecur traverses nodes of tree in preorder */
printPathsRecur(tree, path[], pathlen)
   1) If node is not NULL then 
         a) push data to path array: 
                path[pathlen] = node->data.
         b) increment pathlen 
                pathlen++
   2) If node is a leaf node then print the path array.
   3) Else
        a) Call printPathsRecur for left subtree
                 printPathsRecur(node->left, path, pathLen)
        b) Call printPathsRecur for right subtree.
                printPathsRecur(node->right, path, pathLen)

Example:

Output for the above example will be

  1 2 4
  1 2 5
  1 3 

Implementation:

C++

/* C++ program to print all of its
root-to-leaf paths for a tree*/
#include <bits/stdc++.h>
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;
};

void printArray(int [], int);
void printPathsRecur(node*, int [], int);
node* newNode(int );
void printPaths(node*);

/* 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 led 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);
    }
}

/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
node* newNode(int data)
{
    node* Node = new node();
    Node->data = data;
    Node->left = NULL;
    Node->right = NULL;

    return(Node);
}

/* 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 << endl;
}

/* Driver code */
int main()
{
    node *root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);

    /* Print all root-to-leaf
    paths of the input tree */
    printPaths(root);
    return 0;
}

// This code is contributed by rathbhupendra

C

/*program to print all of its root-to-leaf paths for a tree*/
#include <stdio.h>
#include <stdlib.h>

/* 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;
};

void printArray(int [], int);
void printPathsRecur(struct node*, int [], int);
struct node* newNode(int );
void printPaths(struct node*);

/* 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 led 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);
  }
}

/* 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);
}

/* Utility that prints out an array on a line */
void printArray(int ints[], int len)
{
  int i;
  for (i=0; i<len; i++) {
    printf("%d ", ints[i]);
  }
  printf("\n");
}

/* Driver program to test mirror() */
int main()
{
  struct node *root = newNode(1);
  root->left        = newNode(2);
  root->right       = newNode(3);
  root->left->left  = newNode(4);
  root->left->right = newNode(5);

  /* Print all root-to-leaf paths of the input tree */
  printPaths(root);

  getchar();
  return 0;
}

Java

// Java program to print all root to leaf paths

/* 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 led 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 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 all above functions */
    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();

        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);

        /* Print all root-to-leaf paths of the input tree */
        tree.printPaths(tree.root);

    }
}

Python3

# Python3 program to print all of its
# root-to-leaf paths for a tree
class Node:

    # A binary tree node has data,
    # pointer to left child and a
    # pointer to right child
    def __init__(self, data):
        self.data = data
        self.right = None
        self.left = None

def printRoute(stack, root):
    if root == None:
        return

    # append this node to the path array
    stack.append(root.data)
    if(root.left == None and root.right == None):

        # print out all of its
        # root - to - leaf
        print(' '.join([str(i) for i in stack]))

    # otherwise try both subtrees
    printRoute(stack, root.left)
    printRoute(stack, root.right)
    stack.pop()

# Driver Code
root = Node(1);
root.left = Node(2);
root.right = Node(3);
root.left.left = Node(4);
root.left.right = Node(5);
printRoute([], root)

# This code is contributed
# by Farheen Nilofer

C#

using System;

// C# program to print all root to leaf paths

/* 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 led 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 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 all above functions */
    public static void Main(string[] args)
    {
        BinaryTree tree = new BinaryTree();

        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);

        /* Print all root-to-leaf paths of the input tree */
        tree.printPaths(tree.root);

    }
}

// This code is contributed by Shrikant13

Javascript

<script>

    // JavaScript program to print all root to leaf paths

    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

    let 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.*/
    function printPaths(node)
    {
        let path = new Array(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. */
    function printPathsRecur(node, path, 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 led 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 that prints out an array on a line */
    function printArray(ints, len)
    {
        let i;
        for (i = 0; i < len; i++)
            document.write(ints[i] + " ");
        document.write("</br>");
    }

    root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);

    /* Print all root-to-leaf paths of the input tree */
    printPaths(root);

</script>

Output:

1 2 4
1 2 5
1 3

Time Complexity: O(n)
The time complexity of the above algorithm is O(n) where n is the number of nodes in the binary tree. Since we traverse the node once, the time complexity is O(n).

Space Complexity: O(h)
The space complexity of the above algorithm is O(h) where h is the height of the Binary Tree. We are using the recursion stack to keep the path information so the space complexity is O(h).