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

For the below example tree, all root-to-leaf paths are:
10 –> 8 –> 3
10 –> 8 –> 5
10 –> 2 –> 2

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 <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;  
};  
  
/* Prototypes for funtions 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 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 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<<endl;  
}  
  
/* utility 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);  
}  
  
/* 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<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; 
}; 
  
/* Prototypes for funtions 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 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 FUNCTIONS */
/* 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"); 
}     
  
/* utility 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() 
{  
   
  /* 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 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 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 Insorder 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 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 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 Insorder traversal */
        tree.printPaths(tree.root); 
    } 
} 
  
// This code is contributed by Shrikant13 

Output :

10 8 3 
10 8 5 
10 2 2

Time Complexity: O(n²) where n is number of nodes.

References:
http://cslibrary.stanford.edu/110/BinaryTrees.html

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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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python3

C#

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