Given a tree and a node data, your task to reverse the path till that particular Node.
Examples:
Input : 7 / 6 5 / / 4 3 2 1 Data = 4 Output : Inorder of tree 7 6 3 4 2 5 1 Input : 7 / 6 5 / / 4 3 2 1 Data = 2 Output : Inorder of tree 4 6 3 2 7 5 1
The idea is to use a map to store path level wise.
Find the Node path as well as store in the map
the path is
Replace the position with the map nextPos index value
increment the nextpos index and replace the next value
increment the nextpos index and replace the next value
Let’s understand with the code.
C++
// CPP program to Reverse Tree path #include <bits/stdc++.h> using namespace std; // A Binary Tree Node struct Node { int data; struct Node *left, *right; }; // 'data' is input. We need to reverse path from // root to data. // 'level' is current level. // 'temp' that stores path nodes. // 'nextpos' used to pick next item for reversing. Node* reverseTreePathUtil(Node* root, int data, map< int , int >& temp, int level, int & nextpos) { // return NULL if root NULL if (root == NULL) return NULL; // Final condition // if the node is found then if (data == root->data) { // store the value in it's level temp[level] = root->data; // change the root value with the current // next element of the map root->data = temp[nextpos]; // increment in k for the next element nextpos++; return root; } // store the data in perticular level temp[level] = root->data; // We go to right only when left does not // contain given data. This way we make sure // that correct path node is stored in temp[] Node *left, *right; left = reverseTreePathUtil(root->left, data, temp, level + 1, nextpos); if (left == NULL) right = reverseTreePathUtil(root->right, data, temp, level + 1, nextpos); // If current node is part of the path, // then do reversing. if (left || right) { root->data = temp[nextpos]; nextpos++; return (left ? left : right); } // return NULL if not element found return NULL; } // Reverse Tree path void reverseTreePath(Node* root, int data) { // store per level data map< int , int > temp; // it is for replacing the data int nextpos = 0; // reverse tree path reverseTreePathUtil(root, data, temp, 0, nextpos); } // INORDER void inorder(Node* root) { if (root != NULL) { inorder(root->left); cout << root->data << " " ; inorder(root->right); } } // Utility function to create a new tree node Node* newNode( int data) { Node* temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp; } // Driver program to test above functions int main() { // Let us create binary tree shown in above diagram Node* root = newNode(7); root->left = newNode(6); root->right = newNode(5); root->left->left = newNode(4); root->left->right = newNode(3); root->right->left = newNode(2); root->right->right = newNode(1); /* 7 / 6 5 / / 4 3 2 1 */ int data = 4; // Reverse Tree Path reverseTreePath(root, data); // Traverse inorder inorder(root); return 0; } |
Java
// Java program to Reverse Tree path import java.util.*; class solution { // A Binary Tree Node static class Node { int data; Node left, right; }; //class for int values static class INT { int data; }; // 'data' is input. We need to reverse path from // root to data. // 'level' is current level. // 'temp' that stores path nodes. // 'nextpos' used to pick next item for reversing. static Node reverseTreePathUtil(Node root, int data, Map<Integer, Integer> temp, int level, INT nextpos) { // return null if root null if (root == null ) return null ; // Final condition // if the node is found then if (data == root.data) { // store the value in it's level temp.put(level,root.data); // change the root value with the current // next element of the map root.data = temp.get(nextpos.data); // increment in k for the next element nextpos.data++; return root; } // store the data in perticular level temp.put(level,root.data); // We go to right only when left does not // contain given data. This way we make sure // that correct path node is stored in temp[] Node left, right= null ; left = reverseTreePathUtil(root.left, data, temp, level + 1 , nextpos); if (left == null ) right = reverseTreePathUtil(root.right, data, temp, level + 1 , nextpos); // If current node is part of the path, // then do reversing. if (left!= null || right!= null ) { root.data = temp.get(nextpos.data); nextpos.data++; return (left!= null ? left : right); } // return null if not element found return null ; } // Reverse Tree path static void reverseTreePath(Node root, int data) { // store per level data Map< Integer, Integer> temp= new HashMap< Integer, Integer>(); // it is for replacing the data INT nextpos= new INT(); nextpos.data = 0 ; // reverse tree path reverseTreePathUtil(root, data, temp, 0 , nextpos); } // INORDER static void inorder(Node root) { if (root != null ) { inorder(root.left); System.out.print( root.data + " " ); inorder(root.right); } } // Utility function to create a new tree node static Node newNode( int data) { Node temp = new Node(); temp.data = data; temp.left = temp.right = null ; return temp; } // Driver program to test above functions public static void main(String args[]) { // Let us create binary tree shown in above diagram Node root = newNode( 7 ); root.left = newNode( 6 ); root.right = newNode( 5 ); root.left.left = newNode( 4 ); root.left.right = newNode( 3 ); root.right.left = newNode( 2 ); root.right.right = newNode( 1 ); /* 7 / 6 5 / / 4 3 2 1 */ int data = 4 ; // Reverse Tree Path reverseTreePath(root, data); // Traverse inorder inorder(root); } } //contributed by Arnab Kundu |
C#
// C# program to Reverse Tree path using System; using System.Collections.Generic; class GFG { // A Binary Tree Node public class Node { public int data; public Node left, right; } //class for int values public class INT { public int data; } // 'data' is input. We need to reverse // path from root to data. // 'level' is current level. // 'temp' that stores path nodes. // 'nextpos' used to pick next item for reversing. public static Node reverseTreePathUtil(Node root, int data, IDictionary< int , int > temp, int level, INT nextpos) { // return null if root null if (root == null ) { return null ; } // Final condition // if the node is found then if (data == root.data) { // store the value in it's level temp[level] = root.data; // change the root value with the // current next element of the map root.data = temp[nextpos.data]; // increment in k for the next element nextpos.data++; return root; } // store the data in perticular level temp[level] = root.data; // We go to right only when left does not // contain given data. This way we make sure // that correct path node is stored in temp[] Node left, right = null ; left = reverseTreePathUtil(root.left, data, temp, level + 1, nextpos); if (left == null ) { right = reverseTreePathUtil(root.right, data, temp, level + 1, nextpos); } // If current node is part of the path, // then do reversing. if (left != null || right != null ) { root.data = temp[nextpos.data]; nextpos.data++; return (left != null ? left : right); } // return null if not element found return null ; } // Reverse Tree path public static void reverseTreePath(Node root, int data) { // store per level data IDictionary< int , int > temp = new Dictionary< int , int >(); // it is for replacing the data INT nextpos = new INT(); nextpos.data = 0; // reverse tree path reverseTreePathUtil(root, data, temp, 0, nextpos); } // INORDER public static void inorder(Node root) { if (root != null ) { inorder(root.left); Console.Write(root.data + " " ); inorder(root.right); } } // Utility function to create // a new tree node public static Node newNode( int data) { Node temp = new Node(); temp.data = data; temp.left = temp.right = null ; return temp; } // Driver Code public static void Main( string [] args) { // Let us create binary tree // shown in above diagram Node root = newNode(7); root.left = newNode(6); root.right = newNode(5); root.left.left = newNode(4); root.left.right = newNode(3); root.right.left = newNode(2); root.right.right = newNode(1); /* 7 / 6 5 / / 4 3 2 1 */ int data = 4; // Reverse Tree Path reverseTreePath(root, data); // Traverse inorder inorder(root); } } // This code is contributed by Shrikant13 |
Output:
7 6 3 4 2 5 1
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