Bug Report for serialize-and-deserialize-binary-tree

[kritagyasaxenaa]

Bug Report for https://neetcode.io/problems/serialize-and-deserialize-binary-tree

This is a test case of my code that failed on LeetCode, but my solution was accepted by NeetCode, kindly add this test case, for better accuracy.

The errors in my solution were in toString and toInt, which is incorrectly done. Take it up please.

My solution was:
/**

• Definition for a binary tree node.

• struct TreeNode {

• int val;
  

• TreeNode *left;
  

• TreeNode *right;
  

• TreeNode() : val(0), left(nullptr), right(nullptr) {}
  

• TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
  

• TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
  

• };
  */
  class Codec {
  public:

  vector<TreeNode*>BFS(TreeNode* p){
  vector<TreeNode*>bfs;
  bfs.push_back(p);
  for(int i=0;i<bfs.size();i++){
  if(bfs[i]==nullptr){
  continue;
  }
  else{
  bfs.push_back(bfs[i]->left);
  bfs.push_back(bfs[i]->right);
  }
  }
  return bfs;
  }
  string toString(int a){
  string s;
  while(a!=0){
  int b=a%10;
  char c='0'+ b;
  s+=c;
  a=a/10;
  }
  reverse(begin(s),end(s));
  return s;
  }

  int toInt(string s){
  int a=0;
  for(char i:s){
  a=a*10;
  a+=i-'0';
  }
  return a;
  }

  // Encodes a tree to a single string.
  string serialize(TreeNode* root) {
  // i could simply use bfs
  vector<TreeNode*>bfs=BFS(root);
  string s;
  for(TreeNode* i: bfs){
  if(i==nullptr){
  s.append(",");
  s.append("nullptr");
  }
  else{
  s.append(",");
  s.append(toString(i->val));
  }
  }
  return s;

  }

  // Decodes your encoded data to tree.
  TreeNode* deserialize(string data) {
  if(data.size()==0){
  return nullptr;
  }
  vector<TreeNode*>bfs;
  for(int i=1;i<data.size();i++){
  string b;
  while(i<data.size()&&data[i]!=','){
  b += data[i++];
  }
  if(b=="nullptr"){
  bfs.push_back(nullptr);
  }
  else{
  int a=toInt(b);
  TreeNode* temp=new TreeNode(a);
  bfs.push_back(temp);
  }
  }
  // found the bfs of the tree
  TreeNode* root=bfs[0];
  TreeNode* temp= root;
  for(int i=0,j=1;j<bfs.size();i++){
  if(bfs[i]==nullptr){
  continue;
  }
  bfs[i]->left=bfs[j];
  bfs[i]->right=bfs[j+1];
  j+=2;
  }
  return root;
  }
  };

Regards,
Kritagya