Given a binary tree with distinct nodes(no two nodes have the same data values). The problem is to print the path from root to a given node x. If node x is not present then print “No Path”.
Examples:
Input : 1
/ \
2 3
/ \ / \
4 5 6 7
x = 5
Output : 1->2->5
Approach: Create a recursive function that traverses the different path in the binary tree to find the required node x. If node x is present then it returns true and accumulates the path nodes in some array arr[]. Else it returns false.
Following are the cases during the traversal:
- If root = NULL, return false.
- push the root’s data into arr[].
- if root’s data = x, return true.
- if node x is present in root’s left or right subtree, return true.
- Else remove root’s data value from arr[] and return false.
This recursive function can be accessed from other function to check whether node x is present or not and if it is present, then the path nodes can be accessed from arr[]. You can define arr[] globally or pass its reference to the recursive function.
Implementation:
C++
// C++ implementation to print the path from root// to a given node in a binary tree#include <bits/stdc++.h>using namespace std;// structure of a node of binary treestruct Node{ int data; Node *left, *right;};/* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct Node* getNode(int data){ struct Node *newNode = new Node; newNode->data = data; newNode->left = newNode->right = NULL; return newNode;}// Returns true if there is a path from root// to the given node. It also populates // 'arr' with the given pathbool hasPath(Node *root, vector<int>& arr, int x){ // if root is NULL // there is no path if (!root) return false; // push the node's value in 'arr' arr.push_back(root->data); // if it is the required node // return true if (root->data == x) return true; // else check whether the required node lies // in the left subtree or right subtree of // the current node if (hasPath(root->left, arr, x) || hasPath(root->right, arr, x)) return true; // required node does not lie either in the // left or right subtree of the current node // Thus, remove current node's value from // 'arr'and then return false arr.pop_back(); return false; }// function to print the path from root to the// given node if the node lies in the binary treevoid printPath(Node *root, int x){ // vector to store the path vector<int> arr; // if required node 'x' is present // then print the path if (hasPath(root, arr, x)) { for (int i=0; i<arr.size()-1; i++) cout << arr[i] << "->"; cout << arr[arr.size() - 1]; } // 'x' is not present in the binary tree else cout << "No Path";}// Driver program to test aboveint main(){ // binary tree formation struct Node *root = getNode(1); root->left = getNode(2); root->right = getNode(3); root->left->left = getNode(4); root->left->right = getNode(5); root->right->left = getNode(6); root->right->right = getNode(7); int x = 5; printPath(root, x); return 0;} |
Java
// Java implementation to print the path from root // to a given node in a binary tree import java.util.ArrayList;public class PrintPath { // Returns true if there is a path from root // to the given node. It also populates // 'arr' with the given path public static boolean hasPath(Node root, ArrayList<Integer> arr, int x) { // if root is NULL // there is no path if (root==null) return false; // push the node's value in 'arr' arr.add(root.data); // if it is the required node // return true if (root.data == x) return true; // else check whether the required node lies // in the left subtree or right subtree of // the current node if (hasPath(root.left, arr, x) || hasPath(root.right, arr, x)) return true; // required node does not lie either in the // left or right subtree of the current node // Thus, remove current node's value from // 'arr'and then return false arr.remove(arr.size()-1); return false; } // function to print the path from root to the // given node if the node lies in the binary tree public static void printPath(Node root, int x) { // ArrayList to store the path ArrayList<Integer> arr=new ArrayList<>(); // if required node 'x' is present // then print the path if (hasPath(root, arr, x)) { for (int i=0; i<arr.size()-1; i++) System.out.print(arr.get(i)+"->"); System.out.print(arr.get(arr.size() - 1)); } // 'x' is not present in the binary tree else System.out.print("No Path"); } public static void main(String args[]) { Node 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); root.right.left = new Node(6); root.right.right = new Node(7); int x=5; printPath(root, x); }}// A node of binary tree class Node { int data; Node left, right; Node(int data) { this.data=data; left=right=null; }}; //This code is contributed by Gaurav Tiwari |
Python3
# Python3 implementation to print the path from # root to a given node in a binary tree # Helper Class that allocates a new node # with the given data and None left and # right pointers. class getNode: def __init__(self, data): self.data = data self.left = self.right = None# Returns true if there is a path from # root to the given node. It also # populates 'arr' with the given path def hasPath(root, arr, x): # if root is None there is no path if (not root): return False # push the node's value in 'arr' arr.append(root.data) # if it is the required node # return true if (root.data == x): return True # else check whether the required node # lies in the left subtree or right # subtree of the current node if (hasPath(root.left, arr, x) or hasPath(root.right, arr, x)): return True # required node does not lie either in # the left or right subtree of the current # node. Thus, remove current node's value # from 'arr'and then return false arr.pop(-1) return False# function to print the path from root to # the given node if the node lies in# the binary tree def printPath(root, x): # vector to store the path arr = [] # if required node 'x' is present # then print the path if (hasPath(root, arr, x)): for i in range(len(arr) - 1): print(arr[i], end = "->") print(arr[len(arr) - 1]) # 'x' is not present in the # binary tree else: print("No Path")# Driver Codeif __name__ == '__main__': # binary tree formation root = getNode(1) root.left = getNode(2) root.right = getNode(3) root.left.left = getNode(4) root.left.right = getNode(5) root.right.left = getNode(6) root.right.right = getNode(7) x = 5 printPath(root, x) # This code is contributed by PranchalK |
C#
// C# implementation to print the path from root // to a given node in a binary tree using System;using System.Collections;using System.Collections.Generic;class PrintPath { // A node of binary tree public class Node { public int data; public Node left, right; public Node(int data) { this.data = data; left = right = null; }} // Returns true if there is a path from root // to the given node. It also populates // 'arr' with the given path public static Boolean hasPath(Node root, List<int> arr, int x) { // if root is NULL // there is no path if (root == null) return false; // push the node's value in 'arr' arr.Add(root.data); // if it is the required node // return true if (root.data == x) return true; // else check whether the required node lies // in the left subtree or right subtree of // the current node if (hasPath(root.left, arr, x) || hasPath(root.right, arr, x)) return true; // required node does not lie either in the // left or right subtree of the current node // Thus, remove current node's value from // 'arr'and then return false arr.RemoveAt(arr.Count - 1); return false; } // function to print the path from root to the // given node if the node lies in the binary tree public static void printPath(Node root, int x) { // List to store the path List<int> arr = new List<int>(); // if required node 'x' is present // then print the path if (hasPath(root, arr, x)) { for (int i = 0; i < arr.Count - 1; i++) Console.Write(arr[i]+"->"); Console.Write(arr[arr.Count - 1]); } // 'x' is not present in the binary tree else Console.Write("No Path"); } // Driver code public static void Main(String []args) { Node 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); root.right.left = new Node(6); root.right.right = new Node(7); int x=5; printPath(root, x); }}// This code is contributed by Arnab Kundu |
Javascript
<script>// Javascript implementation to print// the path from root to a given node// in a binary treeclass Node{ constructor(data) { this.left = null; this.right = null; this.data = data; }}// Returns true if there is a path from root // to the given node. It also populates // 'arr' with the given path function hasPath(root, arr, x) { // If root is NULL // there is no path if (root == null) return false; // Push the node's value in 'arr' arr.push(root.data); // If it is the required node // return true if (root.data == x) return true; // Else check whether the required node lies // in the left subtree or right subtree of // the current node if (hasPath(root.left, arr, x) || hasPath(root.right, arr, x)) return true; // Required node does not lie either in the // left or right subtree of the current node // Thus, remove current node's value from // 'arr'and then return false arr.pop(); return false; } // Function to print the path from root to the // given node if the node lies in the binary tree function printPath(root, x) { // ArrayList to store the path let arr = []; // If required node 'x' is present // then print the path if (hasPath(root, arr, x)) { for(let i = 0; i < arr.length - 1; i++) document.write(arr[i] + "->"); document.write(arr[arr.length - 1]); } // 'x' is not present in the binary tree else document.write("No Path"); } // Driver codelet 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); root.right.left = new Node(6); root.right.right = new Node(7); let x = 5;printPath(root, x);// This code is contributed by divyeshrabadiya07</script> |
Output
1->2->5
Time complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(H) where H = height of the binary tree.
Another Approach(Iterative Approach Using Stack):
Follow the below steps to solve the above problem:
1) Start at the root node and push it onto a stack.
2) Create a separate stack to store the path from the root to the current node.
3) While the stack is not empty, do the following:
a) Pop the top node from the stack and add it to the path stack.
b) If the current node is the target node, print the nodes in the path stack to get the path from the root to the target node.
c) Push the right child of the current node onto the stack if it exists.
d) Push the left child of the current node onto the stack if it exists.
Below is the implementation of above approach:
C++
// C++ Program to print the path from root// to a given node in binary tree#include <bits/stdc++.h>using namespace std;// Structure of binary tree nodestruct Node { int data; Node* left; Node* right; Node(int value){ data = value; left = right = NULL; }};// function which will print the pathvoid printPath(Node* root, int target) { vector<int> path; stack<Node*> nodeStack; Node* curr = root; Node* prev = NULL; while (curr || !nodeStack.empty()){ while (curr){ nodeStack.push(curr); path.push_back(curr->data); curr = curr->left; } curr = nodeStack.top(); if (curr->right && curr->right != prev){ curr = curr->right; }else{ if (curr->data == target){ for(int i = 0; i<path.size()-1; i++) cout<<path[i]<<"->"; cout<<path[path.size()-1]<<endl; return; } nodeStack.pop(); path.pop_back(); prev = curr; curr = NULL; } } cout<<"No Path"<<endl;}// Driver program to test above functionsint main() { // Create a binary tree Node *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); root->right->left = new Node(6); root->right->right = new Node(7); int target = 5; printPath(root, target); return 0;}// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002) |
Output
1->2->5
Time Complexity: O(N), where N is the number of nodes in given binary tree.
Auxiliary Space: O(H), where H is the height of the binary tree.
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