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

URI Online Judge Solution URI Solution - Graph

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 existen", s[2]);
else printf("%c nao existen", 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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