# Description

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.

The original problem is here.

The original code is here.

# My Solution

I solve this problem in C++, as below:

``````/*
*Binary Search Tree Iterator
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<stack>
#include<stdlib.h>
using namespace std;

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class BSTIterator {
public:
int nextMin = 0;
stack<TreeNode*> myStack;
BSTIterator(TreeNode *root) {
while(root!=NULL){
myStack.push(root);
root=root->left;
}
}

/** @return whether we have a next smallest number */
bool hasNext() {
if(!myStack.empty()){
TreeNode * node = myStack.top();
myStack.pop();
nextMin = node->val;

node = node->right;
while(node){
myStack.push(node);
node=node->left;
}

return true;
}
return false;
}

/** @return the next smallest number */
int next() {
return nextMin;
}
};

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

# Note

To get the next smallest value in the binary tree, is similar to traversal the tree by in-order.