Mirror of a Tree: Mirror of a Binary Tree T is another Binary Tree M(T) with left and right children of all non-leaf nodes interchanged.
Trees in the above figure are mirror of each other
Method 1 (Recursive)
Algorithm – Mirror(tree):
(1) Call Mirror for left-subtree i.e., Mirror(left-subtree)
(2) Call Mirror for right-subtree i.e., Mirror(right-subtree)
(3) Swap left and right subtrees.
temp = left-subtree
left-subtree = right-subtree
right-subtree = temp
C++
// C++ program to convert a binary tree // to its mirror #include<bits/stdc++.h> using namespace std; /* 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; }; /* Helper function 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); } /* Change a tree so that the roles of the left and right pointers are swapped at every node. So the tree... 4 / 2 5 / 1 3 is changed to... 4 / 5 2 / 3 1 */void mirror(struct Node* node) { if (node == NULL) return; else { struct Node* temp; /* do the subtrees */ mirror(node->left); mirror(node->right); /* swap the pointers in this node */ temp = node->left; node->left = node->right; node->right = temp; } } /* Helper function to print Inorder traversal.*/void inOrder(struct Node* node) { if (node == NULL) return; inOrder(node->left); cout << node->data << " "; inOrder(node->right); } // Driver Code int main() { struct Node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); /* Print inorder traversal of the input tree */ cout << "Inorder traversal of the constructed" << " tree is" << endl; inOrder(root); /* Convert tree to its mirror */ mirror(root); /* Print inorder traversal of the mirror tree */ cout << "
Inorder traversal of the mirror tree" << " is
"; inOrder(root); return 0; } // This code is contributed by Akanksha Rai |
C
// C program to convert a binary tree // to its mirror #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; }; /* Helper function 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); } /* Change a tree so that the roles of the left and right pointers are swapped at every node. So the tree... 4 / 2 5 / 1 3 is changed to... 4 / 5 2 / 3 1 */void mirror(struct Node* node) { if (node==NULL) return; else { struct Node* temp; /* do the subtrees */ mirror(node->left); mirror(node->right); /* swap the pointers in this node */ temp = node->left; node->left = node->right; node->right = temp; } } /* Helper function to print Inorder traversal.*/void inOrder(struct Node* node) { if (node == NULL) return; inOrder(node->left); printf("%d ", node->data); inOrder(node->right); } /* Driver program to test mirror() */int main() { struct Node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); /* Print inorder traversal of the input tree */ printf("Inorder traversal of the constructed" " tree is
"); inOrder(root); /* Convert tree to its mirror */ mirror(root); /* Print inorder traversal of the mirror tree */ printf("
Inorder traversal of the mirror tree" " is
"); inOrder(root); return 0; } |
Java
// Java program to convert binary tree into its mirror /* Class containing left and right child of current node and key value*/class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; } } class BinaryTree { Node root; void mirror() { root = mirror(root); } Node mirror(Node node) { if (node == null) return node; /* do the subtrees */ Node left = mirror(node.left); Node right = mirror(node.right); /* swap the left and right pointers */ node.left = right; node.right = left; return node; } void inOrder() { inOrder(root); } /* Helper function to test mirror(). Given a binary search tree, print out its data elements in increasing sorted order.*/ void inOrder(Node node) { if (node == null) return; inOrder(node.left); System.out.print(node.data + " "); inOrder(node.right); } /* testing for example nodes */ public static void main(String args[]) { /* creating a binary tree and entering the nodes */ BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); /* print inorder traversal of the input tree */ System.out.println("Inorder traversal of input tree is :"); tree.inOrder(); System.out.println(""); /* convert tree to its mirror */ tree.mirror(); /* print inorder traversal of the minor tree */ System.out.println("Inorder traversal of binary tree is : "); tree.inOrder(); } } |
Python3
# Python3 program to convert a binary # tree to its mirror # Utility function to create a new # tree node class newNode: def __init__(self,data): self.data = data self.left = self.right = None """ Change a tree so that the roles of the left and right pointers are swapped at every node. So the tree... 4 / 2 5 / 1 3 is changed to... 4 / 5 2 / 3 1 """def mirror(node): if (node == None): return else: temp = node """ do the subtrees """ mirror(node.left) mirror(node.right) """ swap the pointers in this node """ temp = node.left node.left = node.right node.right = temp """ Helper function to print Inorder traversal."""def inOrder(node) : if (node == None): return inOrder(node.left) print(node.data, end = " ") inOrder(node.right) # Driver code if __name__ =="__main__": root = newNode(1) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(4) root.left.right = newNode(5) """ Print inorder traversal of the input tree """ print("Inorder traversal of the", "constructed tree is") inOrder(root) """ Convert tree to its mirror """ mirror(root) """ Print inorder traversal of the mirror tree """ print("
Inorder traversal of", "the mirror treeis ") inOrder(root) # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to convert binary // tree into its mirror using System; // Class containing left and right // child of current node and key value public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; } } class GFG { public Node root; public virtual void mirror() { root = mirror(root); } public virtual Node mirror(Node node) { if (node == null) { return node; } /* do the subtrees */ Node left = mirror(node.left); Node right = mirror(node.right); /* swap the left and right pointers */ node.left = right; node.right = left; return node; } public virtual void inOrder() { inOrder(root); } /* Helper function to test mirror(). Given a binary search tree, print out its data elements in increasing sorted order.*/public virtual void inOrder(Node node) { if (node == null) { return; } inOrder(node.left); Console.Write(node.data + " "); inOrder(node.right); } /* testing for example nodes */public static void Main(string[] args) { /* creating a binary tree and entering the nodes */ GFG tree = new GFG(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); /* print inorder traversal of the input tree */ Console.WriteLine("Inorder traversal " + "of input tree is :"); tree.inOrder(); Console.WriteLine(""); /* convert tree to its mirror */ tree.mirror(); /* print inorder traversal of the minor tree */ Console.WriteLine("Inorder traversal " + "of binary tree is : "); tree.inOrder(); } } // This code is contributed by Shrikant13 |
Output :
Inorder traversal of the constructed tree is 4 2 5 1 3 Inorder traversal of the mirror tree is 3 1 5 2 4
Time & Space Complexities: This program is similar to traversal of tree space and time complexities will be same as Tree traversal (Please see our Tree Traversal post for details)
Method 2 (Iterative)
The idea is to do queue based level order traversal. While doing traversal, swap left and right children of every node.
// Iterative CPP program to convert a Binary // Tree to its mirror #include<bits/stdc++.h> using namespace std; /* 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; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct Node* newNode(int data) { struct Node* node = new Node; node->data = data; node->left = node->right = NULL; return(node); } /* Change a tree so that the roles of the left and right pointers are swapped at every node. So the tree... 4 / 2 5 / 1 3 is changed to... 4 / 5 2 / 3 1 */void mirror(Node* root) { if (root == NULL) return; queue<Node*> q; q.push(root); // Do BFS. While doing BFS, keep swapping // left and right children while (!q.empty()) { // pop top node from queue Node* curr = q.front(); q.pop(); // swap left child with right child swap(curr->left, curr->right); // push left and right children if (curr->left) q.push(curr->left); if (curr->right) q.push(curr->right); } } /* Helper function to print Inorder traversal.*/void inOrder(struct Node* node) { if (node == NULL) return; inOrder(node->left); cout << node->data << " "; inOrder(node->right); } /* Driver program to test mirror() */int main() { struct Node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); /* Print inorder traversal of the input tree */ cout << "
Inorder traversal of the" " constructed tree is
"; inOrder(root); /* Convert tree to its mirror */ mirror(root); /* Print inorder traversal of the mirror tree */ cout << "
Inorder traversal of the " "mirror tree is
"; inOrder(root); return 0; } |
leave a comment
0 Comments