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Convert a given Binary tree to a tree that holds Logical AND property on C++
In this tutorial, we will be discussing a program to convert a given Binary tree to a tree that holds Logical AND property.
For this we will be provided with a binary tree. Our task is to convert it into a tree that holds the logical AND property means that a node has a value of the AND operation of its children nodes. Note that every node can have a value either zero or one.
Example
#include<bits/stdc++.h> using namespace std; //node structure of binary tree struct Node{ int data; struct Node* left; struct Node* right; }; //creation of a new node struct Node* newNode(int key){ struct Node* node = new Node; node->data= key; node->left = node->right = NULL; return node; } //converting the tree with nodes following //logical AND operation void transform_tree(Node *root){ if (root == NULL) return; //moving to first left node transform_tree(root->left); //moving to first right node transform_tree(root->right); if (root->left != NULL && root->right != NULL) root->data = (root->left->data) & (root->right->data); } //printing the inorder traversal void print_tree(Node* root){ if (root == NULL) return; print_tree(root->left); printf("%d ", root->data); print_tree(root->right); } int main(){ Node *root=newNode(0); root->left=newNode(1); root->right=newNode(0); root->left->left=newNode(0); root->left->right=newNode(1); root->right->left=newNode(1); root->right->right=newNode(1); printf("Before conversion :\n"); print_tree(root); transform_tree(root); printf("\nAfter conversion :\n"); print_tree(root); return 0; }
Output
Before conversion : 0 1 1 0 1 0 1 After conversion : 0 0 1 0 1 1 1
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