URI Online Judge Solution 1200 BST Operations I - Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1200 BST Operations I Tree Recovery  - Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1200 BST Operations I   | Beginner
URI Problem Link - https://www.urionlinejudge.com.br/judge/en/problems/view/1200

Problem Name: 1200 BST Operations I code
Problem Number : URI - 1200 BST Operations I solution
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI Solution 1200 BST Operations I Simple Input output

Sample Input	Sample Output
I c I f I a I h INFIXA PREFIXA POSFIXA P z P h I g INFIXA	a c f h c a f h a h f c z nao existe h existe a c f g h

URI Solution 1200 BST Operations I Code in C++ / URI 1200 in cpp:

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

bool b;
 
struct Node
{
 char data;
 Node* left;
 Node* right;
};
 
Node* GetNewroot(int data)
{
 Node* newroot = new Node();
 newroot -> data = data;
 newroot -> left = NULL;
 newroot -> right = NULL;
 return newroot;
}
 
Node* Insert(Node* root, int data)
{
 if(root == NULL){
  root = GetNewroot(data);
  return root;
 }else if(data <= root -> data){
  root -> left = Insert(root -> left, data);
 }else{
  root -> right = Insert(root -> right, data);
 }
 
 return root;
}

bool lookup(struct Node* root, int target) 
{ 
   if (root == NULL){ 
  return false; 
   }else { 
  if (target == root->data){
   return true; 
  }else{ 
   if (target < root->data) return lookup(root->left, target); 
     else return lookup(root->right, target); 
  } 
   } 
} 
 
void printPreOrder(struct Node* root) 
{
 if (root == NULL) 
  return;
 if(b){
  printf("%c", root -> data);
  b = false; 
 }else{
  printf(" %c", root -> data);
 }
 
 printPreOrder (root -> left); 
 printPreOrder (root -> right); 
}
 
void printInOrder(struct Node* root) 
{ 
 if (root == NULL) 
  return;
 printInOrder (root -> left); 
 if(b){
  printf("%c", root -> data);
  b = false; 
 }else{
  printf(" %c", root -> data);
 }
 printInOrder (root -> right);
}
 
void printPosOrder(struct Node* root) 
{ 
 if (root == NULL) 
  return;
 printPosOrder (root -> left); 
 printPosOrder (root -> right);
 if(b){
  printf("%c", root -> data);
  b = false; 
 }else{
  printf(" %c", root -> data);
 }
}  
 
int main(int argc, char const *argv[])
{
 string s;

 Node* root = NULL;

 while(getline(cin, s))
 {
  if(s == "INFIXA"){
   b = true;
   printInOrder(root);
   printf("\n");
  }else if(s == "PREFIXA"){
   b = true;
   printPreOrder(root);
   printf("\n");
  }else if(s == "POSFIXA"){
   b = true;
   printPosOrder(root);
   printf("\n");
  }else if(s[0] == 'P' && s[1] == ' '){
   if(lookup(root, s[2])) printf("%c existe\n", s[2]);
   else printf("%c nao existe\n", s[2]);
  }else{
   root = Insert(root, s[2]);
  }
 }

 return 0;
}

URI Solution 1200 BST Operations I Code / URI 1200 solution in CPP:

URI Solution 1200 BST Operations I Code / URI 1200 solution in Java:

URI Solution 1200 BST Operations I Code / URI 1200 solution in  Python:

URI Solution 1200 BST Operations I Code / URI 1200 solution in  C# (C Sharp):

Demonstration:

Just implement this in coding. Since having any problem just put a comment below. Thanks



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