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

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




<script>
// javascript 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 {
    constructor(val) {
        this.data = val;
        this.left = null;
        this.right = null;
    }
}
 
var 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) {
        var path = Array(1000).fill(0);
        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 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. */
    function printArray(ints , len) {
        var i;
        for (i = 0; i < len; i++) {
            document.write(ints[i] + " ");
        }
        document.write("<br/>");
    }
 
    // driver program to test above functions
        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);
 
        /* Let us test the built tree by printing Inorder traversal */
        printPaths(root);
 
// This code is contributed by gauravrajput1
</script>

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.

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

Another Method 

C++




#include<bits/stdc++.h>
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<vector<int>> &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<vector<int>> 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<vector<int>> 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<vector<int>> final=Paths(root);
   /*printing paths from root to leaf*/
   for(int i=0;i<final.size();i++)
   {
        
       for(int j=0;j<final[i].size();j++)
            cout<<final[i][j]<<" ";
        cout<<endl;
   }
}

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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