What could be a more efficient and/or elegant way to find if two (or 'n') particular values are present in a...












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I have a solution to find if two values are present in a tree.



I would like to know if there is a

[1] More efficient way to do this ?

[2] More elegant way to this ?

[3] I have mentioned, in comments, the sections of code which I feel could be
done in a better way. Would appreciate suggestions on those sections.

[4] How can we generalize the solution if we have to check if some 'n'
values(all of them) are present in tree or not ?



I have a working solution mentioned in the code below but I think the same could be achieved using a more elegant and efficient code.



/*The structure of a BST Node is as follows:
struct Node {
int data;
Node * right, * left;
};*/

void IsPresent(Node* p, int n1, int n2, bool& f1, bool& f2);

bool CheckIfValuesPresentInTree(Node* root, int n1, int n2) {
auto f1 = bool{false};
auto f2 = bool{false};

IsPresent(root, n1, n2, f1, f2);

return f1&&f2;
}

void IsPresent(Node* p, int n1, int n2, bool& f1, bool& f2) {
// Too many arguments passed to the function.
if (!p) {
return;
}

if (p->data == n1) {
f1 = true;
}

if (p->data == n2) {
f2 = true;
}

IsPresent(p->left, n1, n2, f1, f2);
IsPresent(p->right, n1, n2, f1, f2);
// Second recursive call might not be needed if both n1 and n2 are found
// in 1st call.
// We can write something like below : but is there a better way ?
/*
IsPresent(p->left, n1, n2, f1, f2);
if (f1 && f2) {
return;
}
IsPresent(p->right, n1, n2, f1, f2);
*/
}









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    0












    $begingroup$


    I have a solution to find if two values are present in a tree.



    I would like to know if there is a

    [1] More efficient way to do this ?

    [2] More elegant way to this ?

    [3] I have mentioned, in comments, the sections of code which I feel could be
    done in a better way. Would appreciate suggestions on those sections.

    [4] How can we generalize the solution if we have to check if some 'n'
    values(all of them) are present in tree or not ?



    I have a working solution mentioned in the code below but I think the same could be achieved using a more elegant and efficient code.



    /*The structure of a BST Node is as follows:
    struct Node {
    int data;
    Node * right, * left;
    };*/

    void IsPresent(Node* p, int n1, int n2, bool& f1, bool& f2);

    bool CheckIfValuesPresentInTree(Node* root, int n1, int n2) {
    auto f1 = bool{false};
    auto f2 = bool{false};

    IsPresent(root, n1, n2, f1, f2);

    return f1&&f2;
    }

    void IsPresent(Node* p, int n1, int n2, bool& f1, bool& f2) {
    // Too many arguments passed to the function.
    if (!p) {
    return;
    }

    if (p->data == n1) {
    f1 = true;
    }

    if (p->data == n2) {
    f2 = true;
    }

    IsPresent(p->left, n1, n2, f1, f2);
    IsPresent(p->right, n1, n2, f1, f2);
    // Second recursive call might not be needed if both n1 and n2 are found
    // in 1st call.
    // We can write something like below : but is there a better way ?
    /*
    IsPresent(p->left, n1, n2, f1, f2);
    if (f1 && f2) {
    return;
    }
    IsPresent(p->right, n1, n2, f1, f2);
    */
    }









    share|improve this question









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      0












      0








      0





      $begingroup$


      I have a solution to find if two values are present in a tree.



      I would like to know if there is a

      [1] More efficient way to do this ?

      [2] More elegant way to this ?

      [3] I have mentioned, in comments, the sections of code which I feel could be
      done in a better way. Would appreciate suggestions on those sections.

      [4] How can we generalize the solution if we have to check if some 'n'
      values(all of them) are present in tree or not ?



      I have a working solution mentioned in the code below but I think the same could be achieved using a more elegant and efficient code.



      /*The structure of a BST Node is as follows:
      struct Node {
      int data;
      Node * right, * left;
      };*/

      void IsPresent(Node* p, int n1, int n2, bool& f1, bool& f2);

      bool CheckIfValuesPresentInTree(Node* root, int n1, int n2) {
      auto f1 = bool{false};
      auto f2 = bool{false};

      IsPresent(root, n1, n2, f1, f2);

      return f1&&f2;
      }

      void IsPresent(Node* p, int n1, int n2, bool& f1, bool& f2) {
      // Too many arguments passed to the function.
      if (!p) {
      return;
      }

      if (p->data == n1) {
      f1 = true;
      }

      if (p->data == n2) {
      f2 = true;
      }

      IsPresent(p->left, n1, n2, f1, f2);
      IsPresent(p->right, n1, n2, f1, f2);
      // Second recursive call might not be needed if both n1 and n2 are found
      // in 1st call.
      // We can write something like below : but is there a better way ?
      /*
      IsPresent(p->left, n1, n2, f1, f2);
      if (f1 && f2) {
      return;
      }
      IsPresent(p->right, n1, n2, f1, f2);
      */
      }









      share|improve this question









      $endgroup$




      I have a solution to find if two values are present in a tree.



      I would like to know if there is a

      [1] More efficient way to do this ?

      [2] More elegant way to this ?

      [3] I have mentioned, in comments, the sections of code which I feel could be
      done in a better way. Would appreciate suggestions on those sections.

      [4] How can we generalize the solution if we have to check if some 'n'
      values(all of them) are present in tree or not ?



      I have a working solution mentioned in the code below but I think the same could be achieved using a more elegant and efficient code.



      /*The structure of a BST Node is as follows:
      struct Node {
      int data;
      Node * right, * left;
      };*/

      void IsPresent(Node* p, int n1, int n2, bool& f1, bool& f2);

      bool CheckIfValuesPresentInTree(Node* root, int n1, int n2) {
      auto f1 = bool{false};
      auto f2 = bool{false};

      IsPresent(root, n1, n2, f1, f2);

      return f1&&f2;
      }

      void IsPresent(Node* p, int n1, int n2, bool& f1, bool& f2) {
      // Too many arguments passed to the function.
      if (!p) {
      return;
      }

      if (p->data == n1) {
      f1 = true;
      }

      if (p->data == n2) {
      f2 = true;
      }

      IsPresent(p->left, n1, n2, f1, f2);
      IsPresent(p->right, n1, n2, f1, f2);
      // Second recursive call might not be needed if both n1 and n2 are found
      // in 1st call.
      // We can write something like below : but is there a better way ?
      /*
      IsPresent(p->left, n1, n2, f1, f2);
      if (f1 && f2) {
      return;
      }
      IsPresent(p->right, n1, n2, f1, f2);
      */
      }






      c++ c++11 tree






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      asked 1 hour ago









      kapilkapil

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