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  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
     
       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
     
       stack = new Stack
      
       ();        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 = Noneclass 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())