Binary Search Tree Iterator LeetCode Solution | Easy Approach

Binary Search Tree Iterator 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:

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 Solutions

✅Time:O(n)O(n), next(): O(1)O(1), hasNext(): O(1)O(1)
✅Space: O(n)

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);
  }
};

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);
  }
}

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

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