I am writing a binary tree and I am stuck on a search function that takes a value x and determines if it is a leaf in the tree or not.
Here is what I have:
bool search(Node<T>* ¤tNode, const T& x) const
{
/* Base Case */
if ( currentNode != nullptr && (leaf(currentNode) == true) && currentNode->data == x )
{
return true;
}
/* Recursive Case */
else {
if (currentNode != nullptr)
{
search(currentNode->left, x);
search(currentNode->right, x);
}
//If first if condition is not met for all values then value must not exist
return false;
}
bool leaf(Node<T>* currentNode) const
{
if (currentNode != nullptr)
{
return ((currentNode->left == nullptr && currentNode->right == nullptr) ? true : false);
}
else
{
return true;
}
}
The code always returns false, why is the first IF statement never triggered?
EDIT:
Code calling search:
for (int i = 0; i < 101; i++)
if ((t.search(i) == true) ? cout << "LEAF FOUND: " << i << endl : cout << " NOT A LEAF: " << i << endl);
dealing with recursion can be tricky :D.
unless the root node is a leaf your code will always return return false
your code returns true to the function that made the call not all the way back to the original caller ( before recursion)
let's say your tree had three elements and the item was in the left leaf
so in the root node you called yourself with the left node so it returned true to the caller ( the root node) , the root node does nothing with this and just continue executing.
this can be simply solved simply by adding an if statement to check for the returned value of the functions you called
if (currentNode != nullptr)
{
if(search(currentNode->left, x)) return true;
if(search(currentNode->right, x)) return true;
}
return false;
Because you don't return the boolean from the search method back:
if (currentNode != nullptr)
{
search(currentNode->left, x);
search(currentNode->right, x);
}
You always will return false.
Related
I'm having some trouble figuring out an issue in my code. I needed to do a search to find leaf nodes in a binary tree. When I run the debugger, and select this specific function to call (after creating the tree), I run into an infinite loop. Here is the code I wrote:
void BT::countleaves(node* p)
{
int count = 0;
if (isEmpty())
{
cout << " Tree is empty." << endl;
return;
}
while (p >= NULL)
{
inorderTrav(p->left);
inorderTrav(p->right);
if (p->left == NULL && p->right == NULL)
{
count++;
}
else
{
inorderTrav(p->left);
inorderTrav(p->right);
}
}
}
Why while(p >= NULL) not if(p !=null) ?
else
{
inorderTrav(p->left);
inorderTrav(p->right);
}
Well what for going to both left and right since one of the is null anyhow?
My assignment is to find the sum of all nodes on each branch in a binary search tree using recursion, and compare them to a user input value. If the user input value matches a sum of one of the branches, the function should return true.
In other words, the sum of 32+24+21+14=91. The sum of 32+24+28+25=109. The sum of 32+24+28+31=115 etc. I have tried many different methods, but cant seem to figure out how to traverse each branch accurately. So far I have only been able to traverse and find the sum of the left-most branch.
I am using the method of subtracting each node from the user input value. If the value reaches 0 at a Leaf-node, then clearly the user-input matches the node-sum of that branch on the tree.
The particular points of difficulty for me are when the branch diverges, such as at the node [24] and [28]. I clearly am getting something very simple wrong, but I cant figure it out.
Below is the condensed code I've written so far, in the form of two companion methods (also required for the assignment).
public:
bool findBranchSum1(int value) throw (InvalidTreeArgument) {
if (root == nullptr)
throw InvalidTreeArgument();
return(findBranchSum(root, value));
}
private:
bool findBranchSum(NodePtr node, int value) throw (InvalidTreeArgument)
{
bool result = false;
if (root == nullptr)
throw InvalidTreeArgument();
value -= node->getElement(); //subtract current node from user-input value.
cout << "Current Value = " << value << endl; //help track value changes
if (node->getLeftSide() == nullptr && node->getRightSide() == nullptr)
{
if (value == 0)
{
result = true;
return(true);
}
else
return(false);
}
else
{
if (node->getLeftSide() != nullptr)
{
node = node->getLeftSide(); //advance to next Left node
result = findBranchSum(node, value); //recursive call using new node
}
if (node->getRightSide() != nullptr)
{
node = node->getRightSide(); //advance to next Right node
result = findBranchSum(node, value); //recursive call using new node
}
return(result);
}
}
What am I doing wrong, and how can I fix my code to find the sum of each branch on the tree? Thank you in advance. I apologize for any errors in my format, or missing information.
This is wrong:
if (node->getLeftSide() != nullptr)
{
node = node->getLeftSide(); //advance to next Left node
result = findBranchSum(node, value); //recursive call using new node
}
if (node->getRightSide() != nullptr)
{
node = node->getRightSide(); //advance to next Right node
result = findBranchSum(node, value); //recursive call using new node
}
because you move to the left and then to the right branch of the left (node is changed by your assignment), if it exists! Change to:
if (node->getLeftSide() != nullptr)
{
result = findBranchSum(node->getLeftSide(), value);
}
if (node->getRightSide() != nullptr)
{
result = findBranchSum(node->getRightSide(), value);
}
Your return value management is also broken, change it to:
if (node->getLeftSide() != nullptr)
{
result = findBranchSum(node->getLeftSide(), value);
}
if (!result && node->getRightSide() != nullptr) // cut exploration if previous was correct...
{
result = findBranchSum(node->getRightSide(), value);
}
return result;
if you need to stop at the first correct branch.
I might try something like the following.
bool IsLeaf(Node const * node) {
return node && !node->left && !node->right;
}
bool CheckPathSum(Node const * node, int const target, int const sum_so_far) {
if (!node) return false;
int const sum = sum_so_far + node->element;
if IsLeaf(node) && (sum == target) return true;
return CheckPathSum(node->left, target, sum) ||
CheckPathSum(node->right, target, sum);
}
Call as
CheckPathSum(root, target, 0);
In Java, i tried this-
private static void branchSumsUtil(TreeNode root, List<Integer> sumArray, int runningSum) {
if (root == null){
return;
}
int newRunningSum = runningSum + root.key;
if (root.left == null && root.right == null){
sumArray.add(newRunningSum);
}
branchSumsUtil(root.left, sumArray, newRunningSum);
branchSumsUtil(root.right, sumArray, newRunningSum);
}
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;
}
}
I am writing a binary search tree and this function called Search takes a value x and searches the nodes in the tree and returns whether it is a leaf or not.
bool search(Node<T>* ¤tNode, const T& x) const
{
//~ cout << "CURRENT NODE DATA: " << currentNode->data << " : ";
/* FUNCTION: Searches for variable that is passed in X and checks if this value is a leaf or not */
//Left Subtree Search
if (x < binTree<T>::root->data)
{
if ((leaf(currentNode)) == true)
{
return true;
}
else
{
search(currentNode->left, x);
}
}
//Right Subtree Search
else if (x >= binTree<T>::root->data)
{
//If node in right subtree is a node check
if ((leaf(currentNode)) == true)
{
return true;
}
else
{
search(currentNode->right, x);
}
}
//Return false if the node is not a leaf
return false;
} //END OF SEARCH FUNCTION
bool leaf(Node<T>* currentNode) const
{
return ((currentNode->left == nullptr && currentNode->right == nullptr) ? true : false);
}
The seg fault occurs when I recursively call the search function with new updated node. The binary tree is initialized with 100 values and it starts searching at the root.
This code
if (x < binTree<T>::root->data)
is checking against root, note currentNode, so the test will never change. So if your x value is less than root->data you will keep trying to recurse through currentNode->left until you either hit a leaf (if you are lucky) or you hit a node with a NULL left pointer, in which case you will recurse with currentNode as NULL, which will cause a segfault in leaf when it tries to check for currentNode->left
You should be checking against currentNode. You should also be returning the return value of your search recursive calls
You have to declare
bool leaf(Node<T>* currentNode) const
BEFORE search
Easy fix, copy and paste bool leaf(Node<T>* currentNode) const; before your search function
I am writing a binary tree with a search function. The function is supposed to take an argument x which denotes the value to be searched and once it is found, determine if it is a leaf or not. If the value is not found, the search functions returns false.
Here is what I have which gives me a seg fault and has been just giving me false returns for every value from 1 to 100. The binary tree is initalized with 100 values.
bool search(Node<T>* ¤tNode, const T& x) const
{
//~ cout << "CURRENT NODE DATA: " << currentNode->data << " : ";
/* FUNCTION: Searches for variable that is passed in X and checks if this value is a leaf or not */
//Left Subtree Search
if (x < currentNode->data)
{
if ((leaf(currentNode)) == true)
{
return true;
}
else
{
search(currentNode->left, x);
}
}
//Right Subtree Search
else if (x >= currentNode->data)
{
//If node in right subtree is a node check
if ((leaf(currentNode)) == true)
{
return true;
}
else
{
search(currentNode->right, x);
}
}
//Return false if the node is not a leaf
return false;
} //END OF SEARCH FUNCTION
void remove(Node<T>* ¤tNode, const T& x)
{
}
bool leaf(Node<T>* currentNode) const
{
if (currentNode != nullptr)
{
return ((currentNode->left == nullptr && currentNode->right == nullptr) ? true : false);
}
else
{
return true;
}
}
Additionally, the leaf function returns true when currentNode is nullptr, assuming the check for nullptr in search is resolved. Is this the behavior you want for a pointer pointing to a non-existent leaf?
In your search function, you recourse on the child nodes without seeing if they're nullptr first... Then you try to de reference the data element still without checking for null.