Binary Search Tree Iterator LeetCode Programming Solutions 2022 | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Binary Search Tree Iterator in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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  • Mathematics/Basic Logical Based Questions
  • Arrays
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  • Dynamic Programming
  • Stack & Queue
  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

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Link for the ProblemBinary Search Tree Iterator– LeetCode Problem

Binary Search Tree Iterator– LeetCode Problem

Problem:

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

  • BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
  • boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
  • int next() Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

Example 1:

bst tree
Input
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output

[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]); bSTIterator.next(); // return 3 bSTIterator.next(); // return 7 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 9 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 15 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 20 bSTIterator.hasNext(); // return False

Constraints:

  • The number of nodes in the tree is in the range [1, 105].
  • 0 <= Node.val <= 106
  • At most 105 calls will be made to hasNext, and next.
Binary Search Tree Iterator– LeetCode Solutions
Binary Search Tree Iterator Solution in C++:
class BSTIterator {
 public:
  BSTIterator(TreeNode* root) {
    inorder(root);
  }

  /** @return the next smallest number */
  int next() {
    return vals[i++];
  }

  /** @return whether we have a next smallest number */
  bool hasNext() {
    return i < vals.size();
  }

 private:
  int i = 0;
  vector<int> vals;

  void inorder(TreeNode* root) {
    if (!root)
      return;

    inorder(root->left);
    vals.push_back(root->val);
    inorder(root->right);
  }
};
Binary Search Tree Iterator Solution in Java:
class BSTIterator {
  public BSTIterator(TreeNode root) {
    inorder(root);
  }

  /** @return the next smallest number */
  public int next() {
    return vals.get(i++);
  }

  /** @return whether we have a next smallest number */
  public boolean hasNext() {
    return i < vals.size();
  }

  private int i = 0;
  private List<Integer> vals = new ArrayList<>();

  private void inorder(TreeNode root) {
    if (root == null)
      return;

    inorder(root.left);
    vals.add(root.val);
    inorder(root.right);
  }
}
Binary Search Tree Iterator Solution in Python:
class BSTIterator:
  def __init__(self, root: Optional[TreeNode]):
    self.stack = []
    self.pushLeftsUntilNone(root)

  def next(self) -> int:
    root = self.stack.pop()
    self.pushLeftsUntilNone(root.right)
    return root.val

  def hasNext(self) -> bool:
    return self.stack

  def pushLeftsUntilNone(self, root: Optional[TreeNode]):
    while root:
      self.stack.append(root)
      root = root.left
  • Time: Constructor: O(n)O(n), next(): O(1)O(1), hasNext(): O(1)O(1)
  • Space: O(n)O(n)

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