Binary Search Tree Iterator

Problem:

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.

Analysis:
3/16/2018 update:
Remember to check cur != null in hasNext(). Stack might be empty but we still have cur.

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

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

    /** @return the next smallest number */
    public int next() {
        while (cur != null) {
            stack.push(cur);
            cur = cur.left;
        }
        TreeNode temp = stack.pop();
        cur = temp.right;
        return temp.val;
    }
}

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

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This problem is the breakdown of  "binary search tree inorder traversal" iterative version. Let's first see the code of the iterative version:

 public class Solution {  
   public List < Integer > inorderTraversal(TreeNode root) {  
     List < Integer > res = new ArrayList < > ();  
     Stack < TreeNode > stack = new Stack < > ();  
     TreeNode curr = root;  
     while (curr != null || !stack.isEmpty()) {  
       while (curr != null) {  
         stack.push(curr);  
         curr = curr.left;  
       }  
       curr = stack.pop();  
       res.add(curr.val);  
       curr = curr.right;  
     }  
     return res;  
   }  
 }  

If we only check the second while loop, we can see that it's the implementation of next() function. And the condition of the first while loop: curr != null || ! stack.isEmpty() is the implementation of hasNext() function.