For the below example tree, all root-to-leaf paths are: 10 –> 8 –> 3 10 –> 8 –> 5 10 –> 2 –> 2
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 rootnode 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 toleaf in a binary tree"""# binary tree node contains data field , # left and right pointerclass 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 treedef printPaths(root): # list to store path path = [] printPathsRec(root, path, 0)# Helper function to print path from root # to leaf in binary treedef 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 storeddef 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: https://www.geeksforgeeks.org/binary-tree-data-structure/
Another Method
C++
#include<bits/stdc++.h>using namespace std;/*Binary Tree representation using structure where data is in integer and 2 pointerstruct Node* left and struct Node* right represents left and right pointers of a nodein 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; }} |
Java
//Java code for the above approachimport java.util.ArrayList;import java.util.List;// Binary Tree representation using class where data is in integer and 2 pointers// left and right represents left and right pointers of a node in a tree respectivelyclass Node { int data; Node left; Node right; Node(int x) { data = x; left = null; right = null; }}class GFG { // Recursive helper function which will recursively update our ans list. // it takes root of our tree, arr list and ans list as an argument public static void helper(Node root, List<Integer> arr, List<List<Integer>> ans) { if (root == null) return; arr.add(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 list by inserting list arr which have one unique path from root to leaf */ ans.add(new ArrayList<Integer>(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 list simultaneously */ helper(root.left, new ArrayList<Integer>(arr), ans); helper(root.right, new ArrayList<Integer>(arr), ans); } public static List<List<Integer>> Paths(Node root) { /* * creating 2-d list in which each element is a 1-d list having one unique path * from root to leaf */ List<List<Integer>> ans = new ArrayList<List<Integer>>(); /* if root is null then there is no further action require so return */ if (root == null) return ans; List<Integer> arr = new ArrayList<Integer>(); /* * arr is a list 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 list updated with paths so we will return ans * list */ return ans; } public static void main(String[] args) { /* 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 list */ List<List<Integer>> fina = Paths(root); /* printing paths from root to leaf */ for (List<Integer> path : fina) { for (int i : path) { System.out.print(i + " "); } System.out.println(); } }} |
Python3
"""Python program to print all path from root toleaf in a binary tree"""# binary tree node contains data field ,# left and right pointerclass 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 argumentdef 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 storeddef 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 |
C#
// C# Program for the above approachusing System;using System.Collections;using System.Collections.Generic;// a binary tree node has data, pointer to left child// and a pointer to right childpublic class Node{ public int data; public Node left, right; public Node(int item){ data = item; left = right = null; }}public class BinaryTree{ public Node root; // Recursive helper function which will recursively update our ans vector. // it takes root of our tree ,arr vector and ans vector as an argument public static void helper(Node root, List<int> arr, List<List<int>> ans){ if(root == null) return; List<int> arr1 = new List<int>(arr); arr1.Add(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.Add(arr1); // after that we will return since we don't 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, arr1, ans); helper(root.right, arr1, ans); } public static List<List<int>> Paths(Node root){ // creating 2d vector in which each element is a 1d vector // having one unique path from root to leaf List<List<int>> ans = new List<List<int>>(); // if root is null then there is no further action require so return if(root == null) return ans; List<int> arr = new List<int>(); // 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 return ans; } // 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); // the answer returned will be stored in final 2-d vector List<List<int>> final = new List<List<int>>(); final = Paths(tree.root); // printing paths from root to leaf for(var i = 0; i<final.Count; i++){ for(var j = 0; j<final[i].Count; j++){ Console.Write(final[i][j] + " "); } Console.WriteLine(""); } }}// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002) |
Javascript
// JavaScript program for above approachclass Node { // constructor to create tree node constructor(data) { this.data = data; this.left = null; this.right = null; }}// Recursive helper function which will recursively update our ans array.// it takes root of our tree ,arr array and ans array as an argumentfunction helper(root, arr, ans) { if (!root) { return; } arr.push(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 array by inserting // array arr which have one unique path from root to leaf ans.push(arr.slice()); arr.pop(); // 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); arr.pop();}function Paths(root) { // creating answer in which each element is a array // having one unique path from root to leaf let ans = []; // if root is null then there is no further action require so return if (!root) { return [[]]; } let 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 storedfunction printArray(paths) { for (let path of paths) { for (let i of path) { console.log(i + ' '); } console.log(); }}// Driver program to test above function/*Constructed binary tree is 10 / \ 8 2 / \ / 3 5 2*/let 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 paths = Paths(root);printArray(paths);// This code is contributed by adityamaharshi21 |
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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