c++ my binary tree is well formed [duplicate] - c++

Question

This question already has answers here:
How do you validate a binary search tree?
(33 answers)
Closed 8 years ago.
i propose a recursive implementation for checking whether binary search tree is valid:
/*
Return true if binary tree is a binary search tree
*/
bool BinaryTree::isBinarySearchTree(BinaryTree* tree, int& prev)
{
if(!isBinarySearchTree(tree->left, tree->data)) // left
return false;
if(tree->value > prev) // here
return false;
else
prev = tree->value;
return isBinaryTree(tree->right); // right
}
i have big doubt on the second check,
if(tree->value > prev) // here
return false;
whats your favorite c++ implementation for this problem?
EDIT
how would you extend to find larger BST in given tree?

Answer 1

It's amazing how many people get this wrong.
Here's an example of a tree which the naive solution fails to reject:
5
/ \
/ \
4 6
/ \ / \
1 7 1 7
Every invocation of a naive check will succeed, since every parent is between its children. Yet, it is clearly not a well-formed binary search tree.
Here's a quick solution:
bool test(Tree* n,
int min=std::numeric_limits<int>::min(),
int max=std::numeric_limits<int>::max()) {
return !n || (
min < n->data && n->data < max
&& test(n->left, min, n->data)
&& test(n->right, n->data, max));
}
This isn't perfect, because it requires that neither INT_MIN nor INT_MAX be present in the tree. Often, BST nodes are ordered by <= instead of <, and making that change would only reserve one value instead of two. Fixing the whole thing is left as an exercise.
Here's a demonstration of how the naive test gets it wrong:
#include <iostream>
#include <limits>
#define T new_tree
struct Tree{
Tree* left;
int data;
Tree* right;
};
Tree* T(int v) { return new Tree{0, v, 0}; }
Tree* T(Tree* l, int v, Tree* r) { return new Tree{l, v, r}; }
bool naive_test(Tree* n) {
return n == 0 || ((n->left == 0 || n->data > n->left->data)
&& (n->right == 0 || n->data < n->right->data)
&& naive_test(n->left) && naive_test(n->right));
}
bool test(Tree* n,
int min=std::numeric_limits<int>::min(),
int max=std::numeric_limits<int>::max()) {
return !n || (
min < n->data && n->data < max
&& test(n->left, min, n->data)
&& test(n->right, n->data, max));
}
const char* goodbad(bool b) { return b ? "good" : "bad"; }
int main(int argc, char**argv) {
auto t = T( T( T(1),4,T(7)), 5, T(T(1),6,T(7)));
std::cerr << "Naive test says " << goodbad(naive_test(t))
<< "; Test says " << goodbad(test(t)) << std::endl;
return 0;
}

Answer 2

Recursive impelentation:
bool is_bst (node *root, int min = INT_MIN, int max = INT_MAX)
{
if (root)
{
// check min/max constaint
if (root->data <= min || root->data >= max)
return false;
if (root->left != NULL)
{
// check if the left node is bigger
// or if it is not BST tree itself
if (root->data < root->left->data ||
!is_bst (root->left, min, root->data))
return false;
}
if (root->right != NULL)
{
// check if the right node is smaller
// or if it is not BST tree itself
if (root->data > root->right->data ||
!is_bst (root->right, root->data, max))
return false;
}
}
return true;
}
Iterative impelentation
node_type curr = root;
node_type prev = null;
std::stack<node_type> stack;
while (1)
{
if(curr != null)
{
stack.push (curr);
curr = curr->left;
continue;
}
if(stack.empty()) // done
return true;
curr = stack.pop ();
if (prev != null)
{
if(curr->data < prev->data)
return false;
}
prev = curr;
curr = curr->right;
}