Search Binary Tree function not work c++ - c++

and I was trying to implement a binary searching tree:
template <typename T>
bool Tree<T>::search(TreeNode<T> *ptr, const T &key) {
if (ptr == 0) {
cout<<"No such data: "<<key<<" in the tree"<<endl;
return false;
}
else{
if (ptr->data == key) {
cout<<"Find a node whose data is "<<key<<endl;
return true;
}
else if (ptr->data < key) return search(ptr->leftPtr,key);
else return search(ptr->rightPtr,key);
}
}
But the result always returns false no matter the tree contains the key value or not.
Can u guys help me check the code? I tried debug, but still do not know.
Thank you!

Your traversal comparator for left-tree descending is backwards. As such, as soon as you incorrectly descend into the right tree you stand no chance of ever finding that value. Only the root, and root only, will ever be found correctly.
This:
if (ptr->data < key)
return search(ptr->leftPtr,key);
else
return search(ptr->rightPtr,key);
Should read like this:
if (key < ptr->data) // <== note key is LESS THAN node.
return search(ptr->leftPtr,key);
else
return search(ptr->rightPtr,key);
That said, consider this:
template <typename T>
bool Tree<T>::search(TreeNode<T> *ptr, const T &key)
{
if (ptr == 0) {
cout<<"No such data: "<<key<<" in the tree"<<endl;
return false;
}
if (key < ptr->data)
return search(ptr->leftPtr, key);
else if (ptr->data < key)
return search(ptr->rightPtr, key);
cout<<"Found a node whose data is "<< key << endl;
return true;
}

Related

access to an object member passed by unique_ptr

I wrote the following code to check if a node is in the BST:
bool BST_Node :: BST_Find(unique_ptr<BST_Node> root, int key){
if(!root || root->key == INT_MIN) return false;
if(root->key == key) return true;
else if(key < root->key) BST_Find(move(root->left), key);
else BST_Find(move(root->right), key);
}
root parameter is passed using move(bst) where bst in an unique_ptr.
The problem is when it tries to read root->key: even if the key is present in the tree, this method returns false.
I've tried to use the debugger and root can not be accessed.
Here is the code that uses this method:
auto bst = make_unique<BST_Node>();
for(int i=0; i<n; i++){
key = rand();
if(!bst->BST_Find(move(bst), key)) {
bst->BST_Insert(move(bst), key, "");
}
}
Try this
bool BST_Node :: BST_Find(unique_ptr<BST_Node> const &root, int key){
if(!root || root->key == INT_MIN) return false;
if(root->key == key) return true;
else if(key < root->key) return BST_Find(root->left, key);
else return BST_Find(root->right, key);
}
BST_Find(bst, 42) // no move

Find ancestors of given node in a n-ary tree c++

Given a larg n-ary tree, I need to create a recursive function that prints all the ancestors of a leaf for example where the n-ary tree structure is given as
typedef struct sNaryNode
{
int *data;
int nchild;
struct sNaryNode **child;
} NaryNode;
Here is the function I used but that gives a wrong answer:
bool printAncestors(NaryNode *root, int *data)
{
int i=0;
if (root == NULL)
return false;
if (root->data == data)
return true;
do
{
auto b=printAncestors(root->child[i], data);
if(b)
{
cout<<*root->data<<" ";
return true;
}
else
i++;
}
while(i<root->nchild);
}
You're missing a return value at the end, and you may enter the loop and access root->child[i] even if root->nchild is zero.
Both of those will cause undefined behaviour.
I would write this with a for-loop instead:
bool printAncestors(const NaryNode *root, const int *data)
{
if (root == nullptr)
return false;
if (root->data == data)
return true;
for (int i = 0; i < root->nchild; i++)
{
if (printAncestors(root->child[i], data))
{
cout << *root->data << " ";
return true;
}
}
return false;
}

How to make this search function non-recursive?

I am trying to turn this recursive function into a non-recursive one.This is a search function from a binary search tree. I am aware it is natural to make it recursive, but for learning purposes I would like to make it non-recursive. How could I do this? Thanks in advance!
bool Search(BstNode* root, string data) {
if (root == NULL) return false;
else if (root->data == data) return true;
else if (data <= root->data) return Search(root->left, data);
else return Search(root->right, data);
}
Here is mechanical way to make a recursive algorithm non-recursive.
bool Search(BstNode* root, string data) {
if (root == NULL) return false;
else if (root->data == data) return true;
else if (data <= root->data) return Search(root->left, data);
else return Search(root->right, data);
}
Bundle up the state (arguments and local variables):
bool Search(BstNode* root, string data) {
struct State {
BstNode* root;
string data;
};
State state{root, data};
if (state.root == NULL) return false;
else if (state.root->data == state.data) return true;
else if (data <= state.root->data) return Search(state.root->left, state.data);
else return Search(state.root->right, state.data);
}
wrap body in a loop:
bool Search(BstNode* root, string data) {
struct State {
BstNode* root;
string data;
};
State state{root, data};
while(true) {
if (state.root == NULL) return false;
else if (state.root->data == state.data) return true;
else if (data <= state.root->data) return Search(state.root->left, data);
else return Search(state.root->right, data);
}
}
Replace case where you tail-end recurse (return recursive_call) with changing state and continue:
bool Search(BstNode* root, string data) {
struct State {
BstNode* root;
string data;
};
State state{root, data};
while(true) {
if (state.root == NULL) return false;
else if (state.root->data == state.data) return true;
else if (data <= state.root->data) {
state = {state.root->left, state.data};
continue;
} else {
state = {state.root->right, state.data};
continue;
}
}
}
Now, if there are any more recursive calls that are not return recursive_call, add a manual stack of state and push/pop it instead of changing the back. Include the location of the return state as a void** in the code with labels.
This isn't required here, so I won't bother doing it.
You can usually make a recursive function in general iterative by essentially 'putting' the recursive calls onto a stack, and then using
while !stack.is_empty() do stack.pop() kind of thing
as this is essentially what a compiler will do given that recursion doesn't happen at the machine-code level

How to check if a tree is a BST?

I have to check if a tree is a binary search tree. I'm doing this with an inorder traversal with a temporary array that collects the values. I have to check if the array is ascending order and if it is then I return true:
bool myisBST(Node* node, std::vector<int> v);
bool myisBST(Node* node)
{
return myisBST(node, std::vector<int>());
}
bool myisBST(Node* node, std::vector<int> v)
{
if (node)
{
if (node->left)
return myisBST(node->left, v);
v.push_back(node->data);
if (node->right)
return myisBST(node->right, v);
}
return std::is_sorted(v.begin(), v.end());
}
When binary tree is this:
50
/ \
25 75
/ \ / \
1 12 62 -99
As you can see, the -99 makes this not a binary search tree, but it is still returning true. Is there something wrong with my implementation?
Demo
Two problems:
In myisBST, you are passing v by value, not by reference, so when you pass the vector on recursively, the changes that are made to it don't change its value in the calling method. Simply change the function signature to bool myisBST(Node* node, std::vector<int>& v) to fix this.
The value you should be returning is whether the vector is sorted (as you do in the last line of your method), but instead you are returning prematurely by writing return myisBST(node->left, v); and return myisBST(node->right, v);. You're not actually interested in the return values of these methods; you're just using them to fill the vector inorder. Remove the return from both of these lines.
Following these two fixes, your method works.
First of all, you should probably pass the vector by reference or each recursive call will get a copy and thus the original vector will probably be empty.
Second, you don't even need to create the vector first and then do the check, you can just check the BST property at each node, i.e., the root must be bigger than the left child and smaller than the right child, e.g.,
bool isBST(const Node* root, vector<int>* v) {
if (!root) { return true; }
bool leftBST = true;
if (root->left) {
if (root->data > root->left->data) {
leftBST = isBST(root->left, v);
} else {
// the current node violates the BST precondition
return false;
}
}
// push the root
v->push_back(root->data);
// return false if left subtree is not a BST
if (!leftBST) return false;
if (root->right) {
if (root->data < root->right->data) {
// return whether or not the right subtree is a BST
return isBST(root->left, v);
} else {
// the current node violates the BST precondition
return false;
}
}
// everything good, this is a BST
return true;
}
C++ Program to check if tree is BST or not
struct Node
{
int data;
struct Node* left, *right;
};
bool IsBST(Node* ObjNode)
{
bool leftBST = false;
bool rightBST = false;
if( ObjNode->left != null && ObjNode-left < objNode->data)
{
leftBST = IsBST(ObjNode->left)
}
else if( ObjNode->left == null)
{
leftBST = true;
}
else if( ObjNode->left != null && ObjNode-left >= objNode->data)
{
leftBST = false;
}
if( ObjNode->left != null && ObjNode-left < objNode->data)
{
rightBST = IsBST(ObjNode->right)
}
else if( ObjNode->right == null)
{
rightBST = true;
}
else if( ObjNode->right != null && ObjNode-right >= objNode->data)
{
rightBST = false;
}
return (leftBST && rightBST );
}
In the previous solution, they are keeping a list of the inorder traversal, you really don't need it, you can keep checking with the last traversed element and keep moving forward.
Following solution is the fastest
class Solution {
int lastval = Integer.MIN_VALUE;
int count = 0;
public boolean isValidBST(TreeNode root) {
if(root == null) return true;
boolean left = isValidBST(root.left);
if(!left){
return false;
}
int rootVal = root.val;
if(rootVal == -2147483648 && count == 0 ){
rootVal = rootVal + 1;
}
if( rootVal <= lastval){
return false;
}
count ++;
lastval = root.val;
boolean right = isValidBST(root.right);
if(!right){
return false;
}
return true;
}
}

Binary search tree lazy deletion

I want to implement a lazy deletion into my code. I have added a boolean value to keep track of when a node is "deleted"/marked deleted. I am unsure of what methods to actually change. Here are my remove and insert methods. I attempted the remove, but not the insert. I know for sure that the insert method will have to do a lot of checking. Please advise.
template<class Comparable>
bool search_tree<Comparable>::remove(treeNode<Comparable> * &root,
const Comparable &x) {
if (root == NULL)
return false;
if (x < root->data)
return remove(root->lftChild, x);
if (root->data < x)
return remove(root->rtChild, x);
root->deleted = true;
return true;
}
template<class Comparable>
bool search_tree<Comparable>::insert(treeNode<Comparable> * &root,
const Comparable &x) {
if (root == NULL) {
root = new treeNode<Comparable>(x, NULL, NULL);
return true;
} else if (x < root->data)
return insert(root->lftChild, x);
else if (root->data < x)
return insert(root->rtChild, x);
return false;
}
Your insert function should work as is unless x is equal to an item that has been deleted then it wont get inserted. If x is equal to root->data you need to make sure deleted is false.
And your remove function looks like it should work fine.