173. Binary Search Tree Iterator

Problem Description

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.

Examples

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

class BSTIterator:
    def __init__(self, root: Optional[TreeNode]):
        self.stack = []
        self._leftmost_inorder(root)
        
    def _leftmost_inorder(self, root):
        while root:
            self.stack.append(root)
            root = root.left
    
    def next(self) -> int:
        topmost_node = self.stack.pop()
        if topmost_node.right:
            self._leftmost_inorder(topmost_node.right)
        return topmost_node.val
        
    def hasNext(self) -> bool:
        return len(self.stack) > 0

class BSTIterator {
    private Stack stack;
    
    public BSTIterator(TreeNode root) {
        stack = new Stack<>();
        leftmostInorder(root);
    }
    
    private void leftmostInorder(TreeNode root) {
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
    }
    
    public int next() {
        TreeNode topmostNode = stack.pop();
        if (topmostNode.right != null) {
            leftmostInorder(topmostNode.right);
        }
        return topmostNode.val;
    }
    
    public boolean hasNext() {
        return !stack.isEmpty();
    }
}

class BSTIterator {
private:
    stack stack;
    
    void leftmostInorder(TreeNode* root) {
        while (root) {
            stack.push(root);
            root = root->left;
        }
    }
    
public:
    BSTIterator(TreeNode* root) {
        leftmostInorder(root);
    }
    
    int next() {
        TreeNode* topmostNode = stack.top();
        stack.pop();
        if (topmostNode->right) {
            leftmostInorder(topmostNode->right);
        }
        return topmostNode->val;
    }
    
    bool hasNext() {
        return !stack.empty();
    }
};

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
class BSTIterator {
    constructor(root) {
        this.stack = [];
        this.leftmostInorder(root);
    }
    
    leftmostInorder(root) {
        while (root) {
            this.stack.push(root);
            root = root.left;
        }
    }
    
    next() {
        const topmostNode = this.stack.pop();
        if (topmostNode.right) {
            this.leftmostInorder(topmostNode.right);
        }
        return topmostNode.val;
    }
    
    hasNext() {
        return this.stack.length > 0;
    }
}

public class BSTIterator {
    private Stack stack;
    
    public BSTIterator(TreeNode root) {
        stack = new Stack();
        LeftmostInorder(root);
    }
    
    private void LeftmostInorder(TreeNode root) {
        while (root != null) {
            stack.Push(root);
            root = root.left;
        }
    }
    
    public int Next() {
        TreeNode topmostNode = stack.Pop();
        if (topmostNode.right != null) {
            LeftmostInorder(topmostNode.right);
        }
        return topmostNode.val;
    }
    
    public bool HasNext() {
        return stack.Count > 0;
    }
}