Binary Search Tree Iterator

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

Approach #1: C++.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    BSTIterator(TreeNode *root) {
        helper(root);
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        if (minNums.empty()) return false;
        else return true;
    }

    /** @return the next smallest number */
    int next() {
        TreeNode* cur = minNums.top();
        minNums.pop();
        helper(cur->right);
        return cur->val;
    }
private:
    stack<TreeNode*> minNums;
    void helper(TreeNode* root) {
        while (root != nullptr) {
            minNums.push(root);
            root = root->left;
        }
    }
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */

　　

Approach #2: Java.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {

    public BSTIterator(TreeNode root) {
        helper(root);
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode cur = stack.pop();
        helper(cur.right);
        return cur.val;
    }
    
    private Stack<TreeNode> stack = new Stack<TreeNode>();
    
    private void helper(TreeNode root) {
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
    }
    
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */

　　

Approach #3: Python.

# Definition for a  binary tree node
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class BSTIterator(object):
    def __init__(self, root):
        """
        :type root: TreeNode
        """
        self.stack = list()
        self.helper(root)
        

    def hasNext(self):
        """
        :rtype: bool
        """
        return self.stack
        

    def next(self):
        """
        :rtype: int
        """
        cur = self.stack.pop()
        self.helper(cur.right)
        return cur.val
        
    def helper(self, root):
        while root is not None:
            self.stack.append(root)
            root = root.left
        

# Your BSTIterator will be called like this:
# i, v = BSTIterator(root), []
# while i.hasNext(): v.append(i.next())