Binary Search Tree Iterator LeetCode Solution

Problem – 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.

Follow up:

  • Could you implement next() and hasNext() to run in average O(1) time and use O(h) memory, where h is the height of the tree?

Binary Search Tree Iterator LeetCode Solution in Java

public class BSTIterator {
    private Stack<TreeNode> stack = new Stack<TreeNode>();
    
    public BSTIterator(TreeNode root) {
        pushAll(root);
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode tmpNode = stack.pop();
        pushAll(tmpNode.right);
        return tmpNode.val;
    }
    
    private void pushAll(TreeNode node) {
        for (; node != null; stack.push(node), node = node.left);
    }
}

Binary Search Tree Iterator LeetCode Solution in C++

class BSTIterator {
    stack<TreeNode *> myStack;
public:
    BSTIterator(TreeNode *root) {
        pushAll(root);
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !myStack.empty();
    }

    /** @return the next smallest number */
    int next() {
        TreeNode *tmpNode = myStack.top();
        myStack.pop();
        pushAll(tmpNode->right);
        return tmpNode->val;
    }

private:
    void pushAll(TreeNode *node) {
        for (; node != NULL; myStack.push(node), node = node->left);
    }
};

Binary Search Tree Iterator LeetCode Solution in Python

class BSTIterator:
    # @param root, a binary search tree's root node
    def __init__(self, root):
        self.stack = list()
        self.pushAll(root)

    # @return a boolean, whether we have a next smallest number
    def hasNext(self):
        return self.stack

    # @return an integer, the next smallest number
    def next(self):
        tmpNode = self.stack.pop()
        self.pushAll(tmpNode.right)
        return tmpNode.val
        
    def pushAll(self, node):
        while node is not None:
            self.stack.append(node)
            node = node.left
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