What could be a more efficient and/or elegant way to find if two (or 'n') particular values are present in a...
$begingroup$
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);
*/
}
c++ c++11 tree
asked 4 mins ago
kapilkapil
504
$endgroup$
add a comment |
$begingroup$
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);
*/
}
c++ c++11 tree
asked 4 mins ago
kapilkapil
504
$endgroup$
add a comment |
$begingroup$
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);
*/
}
c++ c++11 tree
asked 4 mins ago
kapilkapil
504
$endgroup$
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
c++ c++11 tree
asked 4 mins ago
kapilkapil
504
asked 4 mins ago
kapilkapil
504
asked 4 mins ago
kapilkapil
504
asked 4 mins ago
kapilkapil
504
asked 4 mins ago
kapilkapil
504
504
add a comment |
add a comment |
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