Convert Binary Tree to Doubly Linked List by fixing left and right pointers
Given a Binary Tree (BT), convert it to a Doubly Linked List(DLL). The left and right pointers in nodes are to be used as previous and next pointers respectively in converted DLL. The order of nodes in DLL must be the same as in Inorder for the given Binary Tree. The first node of Inorder traversal (leftmost node in BT) must be the head node of the DLL.
A solution to this problem has been discussed in this post.
In this post, another simple and efficient solution is discussed. The solution discussed here has two simple steps.
- Fix Left Pointers: In this step, we change left pointers to point to previous nodes in DLL. The idea is simple, we do in order traversal of the tree. In order to traversal, we keep track of the previously visited node and change the left pointer to the previous node. See fixPrevPtr() implementation below.
- Fix Right Pointers: The above is intuitive and simple. How to change the right pointers to point to the next node in DLL? The idea is to use left pointers fixed in step 1. We start from the rightmost node in Binary Tree (BT). The rightmost node is the last node in DLL. Since left pointers are changed to point to the previous node in DLL, we can linearly traverse the complete DLL using these pointers. The traversal would be from last to the first node. While traversing the DLL, we keep track of the previously visited node and change the right pointer to the previous node. See fixNextPtr() implementation below.
Implementation:
C++
// A simple inorder traversal based // program to convert a Binary Tree to DLL #include <bits/stdc++.h> using namespace std; // A tree node class node { public : int data; node *left, *right; }; // A utility function to create // a new tree node node *newNode( int data) { node *Node = new node(); Node->data = data; Node->left = Node->right = NULL; return (Node); } // Standard Inorder traversal of tree void inorder(node *root) { if (root != NULL) { inorder(root->left); cout << "\t" << root->data; inorder(root->right); } } // Changes left pointers to work as // previous pointers in converted DLL // The function simply does inorder // traversal of Binary Tree and updates // left pointer using previously visited node void fixPrevPtr(node *root) { static node *pre = NULL; if (root != NULL) { fixPrevPtr(root->left); root->left = pre; pre = root; fixPrevPtr(root->right); } } // Changes right pointers to work // as next pointers in converted DLL node *fixNextPtr(node *root) { node *prev = NULL; // Find the right most node // in BT or last node in DLL while (root && root->right != NULL) root = root->right; // Start from the rightmost node, // traverse back using left pointers. // While traversing, change right pointer of nodes. while (root && root->left != NULL) { prev = root; root = root->left; root->right = prev; } // The leftmost node is head // of linked list, return it return (root); } // The main function that converts // BST to DLL and returns head of DLL node *BTToDLL(node *root) { // Set the previous pointer fixPrevPtr(root); // Set the next pointer and return head of DLL return fixNextPtr(root); } // Traverses the DLL from left to right void printList(node *root) { while (root != NULL) { cout<< "\t" <<root->data; root = root->right; } } // Driver code int main( void ) { // Let us create the tree // shown in above diagram node *root = newNode(10); root->left = newNode(12); root->right = newNode(15); root->left->left = newNode(25); root->left->right = newNode(30); root->right->left = newNode(36); cout<< "\n\t\tInorder Tree Traversal\n\n" ; inorder(root); node *head = BTToDLL(root); cout << "\n\n\t\tDLL Traversal\n\n" ; printList(head); return 0; } // This code is contributed by rathbhupendra |
C
// A simple inorder traversal based program to convert a Binary Tree to DLL #include<stdio.h> #include<stdlib.h> // A tree node struct node { int data; struct node *left, *right; }; // A utility function to create a new tree node struct node *newNode( int data) { struct node *node = ( struct node *) malloc ( sizeof ( struct node)); node->data = data; node->left = node->right = NULL; return (node); } // Standard Inorder traversal of tree void inorder( struct node *root) { if (root != NULL) { inorder(root->left); printf ( "\t%d" ,root->data); inorder(root->right); } } // Changes left pointers to work as previous pointers in converted DLL // The function simply does inorder traversal of Binary Tree and updates // left pointer using previously visited node void fixPrevPtr( struct node *root) { static struct node *pre = NULL; if (root != NULL) { fixPrevPtr(root->left); root->left = pre; pre = root; fixPrevPtr(root->right); } } // Changes right pointers to work as next pointers in converted DLL struct node *fixNextPtr( struct node *root) { struct node *prev = NULL; // Find the right most node in BT or last node in DLL while (root && root->right != NULL) root = root->right; // Start from the rightmost node, traverse back using left pointers. // While traversing, change right pointer of nodes. while (root && root->left != NULL) { prev = root; root = root->left; root->right = prev; } // The leftmost node is head of linked list, return it return (root); } // The main function that converts BST to DLL and returns head of DLL struct node *BTToDLL( struct node *root) { // Set the previous pointer fixPrevPtr(root); // Set the next pointer and return head of DLL return fixNextPtr(root); } // Traverses the DLL from left to right void printList( struct node *root) { while (root != NULL) { printf ( "\t%d" , root->data); root = root->right; } } // Driver program to test above functions int main( void ) { // Let us create the tree shown in above diagram struct node *root = newNode(10); root->left = newNode(12); root->right = newNode(15); root->left->left = newNode(25); root->left->right = newNode(30); root->right->left = newNode(36); printf ( "\n\t\tInorder Tree Traversal\n\n" ); inorder(root); struct node *head = BTToDLL(root); printf ( "\n\n\t\tDLL Traversal\n\n" ); printList(head); return 0; } |
Java
// Java program to convert BTT to DLL using // simple inorder traversal public class BinaryTreeToDLL { static class node { int data; node left, right; public node( int data) { this .data = data; } } static node prev; // Changes left pointers to work as previous // pointers in converted DLL The function // simply does inorder traversal of Binary // Tree and updates left pointer using // previously visited node static void fixPrevptr(node root) { if (root == null ) return ; fixPrevptr(root.left); root.left = prev; prev = root; fixPrevptr(root.right); } // Changes right pointers to work // as next pointers in converted DLL static node fixNextptr(node root) { // Find the right most node in // BT or last node in DLL while (root.right != null ) root = root.right; // Start from the rightmost node, traverse // back using left pointers. While traversing, // change right pointer of nodes while (root != null && root.left != null ) { node left = root.left; left.right = root; root = root.left; } // The leftmost node is head of linked list, return it return root; } static node BTTtoDLL(node root) { prev = null ; // Set the previous pointer fixPrevptr(root); // Set the next pointer and return head of DLL return fixNextptr(root); } // Traverses the DLL from left to right static void printlist(node root) { while (root != null ) { System.out.print(root.data + " " ); root = root.right; } } // Standard Inorder traversal of tree static void inorder(node root) { if (root == null ) return ; inorder(root.left); System.out.print(root.data + " " ); inorder(root.right); } public static void main(String[] args) { // Let us create the tree shown in above diagram node root = new node( 10 ); root.left = new node( 12 ); root.right = new node( 15 ); root.left.left = new node( 25 ); root.left.right = new node( 30 ); root.right.left = new node( 36 ); System.out.println( "Inorder Tree Traversal" ); inorder(root); node head = BTTtoDLL(root); System.out.println( "\nDLL Traversal" ); printlist(head); } } // This code is contributed by Rishabh Mahrsee |
Python3
# A simple inorder traversal based program to convert a # Binary Tree to DLL # A Binary Tree node class Node: # Constructor to create a new tree node def __init__( self , data): self .data = data self .left = None self .right = None # Standard Inorder traversal of tree def inorder(root): if root is not None : inorder(root.left) print ( "\t%d" % (root.data),end = " " ) inorder(root.right) # Changes left pointers to work as previous pointers # in converted DLL # The function simply does inorder traversal of # Binary Tree and updates # left pointer using previously visited node def fixPrevPtr(root): if root is not None : fixPrevPtr(root.left) root.left = fixPrevPtr.pre fixPrevPtr.pre = root fixPrevPtr(root.right) # Changes right pointers to work as next pointers in # converted DLL def fixNextPtr(root): prev = None # Find the right most node in BT or last node in DLL while (root and root.right ! = None ): root = root.right # Start from the rightmost node, traverse back using # left pointers # While traversing, change right pointer of nodes while (root and root.left ! = None ): prev = root root = root.left root.right = prev # The leftmost node is head of linked list, return it return root # The main function that converts BST to DLL and returns # head of DLL def BTToDLL(root): # Set the previous pointer fixPrevPtr(root) # Set the next pointer and return head of DLL return fixNextPtr(root) # Traverses the DLL from left to right def printList(root): while (root ! = None ): print ( "\t%d" % (root.data),end = " " ) root = root.right # Driver program to test above function root = Node( 10 ) root.left = Node( 12 ) root.right = Node( 15 ) root.left.left = Node( 25 ) root.left.right = Node( 30 ) root.right.left = Node( 36 ) print ( "Inorder Tree Traversal" ) inorder(root) # Static variable pre for function fixPrevPtr fixPrevPtr.pre = None head = BTToDLL(root) print ( "\nDLL Traversal" ) printList(head) # This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
// C# program to convert BTT to DLL using // simple inorder traversal using System; class GFG { public class node { public int data; public node left, right; public node( int data) { this .data = data; } } public static node prev; // Changes left pointers to work as previous // pointers in converted DLL The function // simply does inorder traversal of Binary // Tree and updates left pointer using // previously visited node public static void fixPrevptr(node root) { if (root == null ) { return ; } fixPrevptr(root.left); root.left = prev; prev = root; fixPrevptr(root.right); } // Changes right pointers to work // as next pointers in converted DLL public static node fixNextptr(node root) { // Find the right most node in // BT or last node in DLL while (root.right != null ) { root = root.right; } // Start from the rightmost node, traverse // back using left pointers. While traversing, // change right pointer of nodes while (root != null && root.left != null ) { node left = root.left; left.right = root; root = root.left; } // The leftmost node is head of // linked list, return it return root; } public static node BTTtoDLL(node root) { prev = null ; // Set the previous pointer fixPrevptr(root); // Set the next pointer and // return head of DLL return fixNextptr(root); } // Traverses the DLL from left to right public static void printlist(node root) { while (root != null ) { Console.Write(root.data + " " ); root = root.right; } } // Standard Inorder traversal of tree public static void inorder(node root) { if (root == null ) { return ; } inorder(root.left); Console.Write(root.data + " " ); inorder(root.right); } public static void Main() { // Let us create the tree // shown in above diagram node root = new node(10); root.left = new node(12); root.right = new node(15); root.left.left = new node(25); root.left.right = new node(30); root.right.left = new node(36); Console.WriteLine( "Inorder Tree Traversal" ); inorder(root); node head = BTTtoDLL(root); Console.WriteLine( "\nDLL Traversal" ); printlist(head); } } // This code is contributed by Shrikant13 |
Javascript
<script> // javascript program to convert BTT to DLL using // simple inorder traversal class node { constructor(val) { this .data = val; this .left = null ; this .right = null ; } } var prev; // Changes left pointers to work as previous // pointers in converted DLL The function // simply does inorder traversal of Binary // Tree and updates left pointer using // previously visited node function fixPrevptr( root) { if (root == null ) return ; fixPrevptr(root.left); root.left = prev; prev = root; fixPrevptr(root.right); } // Changes right pointers to work // as next pointers in converted DLL function fixNextptr( root) { // Find the right most node in // BT or last node in DLL while (root.right != null ) root = root.right; // Start from the rightmost node, traverse // back using left pointers. While traversing, // change right pointer of nodes while (root != null && root.left != null ) { var left = root.left; left.right = root; root = root.left; } // The leftmost node is head of linked list, return it return root; } function BTTtoDLL( root) { prev = null ; // Set the previous pointer fixPrevptr(root); // Set the next pointer and return head of DLL return fixNextptr(root); } // Traverses the DLL from left to right function printlist( root) { while (root != null ) { document.write(root.data + " " ); root = root.right; } } // Standard Inorder traversal of tree function inorder( root) { if (root == null ) return ; inorder(root.left); document.write(root.data + " " ); inorder(root.right); } // Let us create the tree shown in above diagram var root = new node(10); root.left = new node(12); root.right = new node(15); root.left.left = new node(25); root.left.right = new node(30); root.right.left = new node(36); document.write( "<br/>Inorder Tree Traversal<br/>" ); inorder(root); var head = BTTtoDLL(root); document.write( "<br/>DLL Traversal<br/>" ); printlist(head); // This code contributed by umadevi9616 </script> |
Inorder Tree Traversal 25 12 30 10 36 15 DLL Traversal 25 12 30 10 36 15
Time Complexity: O(n) where n is the number of nodes in a given Binary Tree. The solution simply does two traversals of all Binary Tree nodes.
Auxiliary Space: O(n)
The extra space is used due to recursion call stack of fixPrevPtr function. The worst case happens when the tree is skewed.
Please Login to comment...