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.

Solution

# 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 = []
        self.cur = root

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

    def next(self):
        """
        :rtype: int
        """
        while self.cur:
            self.stack.append(self.cur)
            self.cur = self.cur.left
        self.cur = self.stack.pop()
        node = self.cur
        self.cur = self.cur.right

        return node.val


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

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

public class BSTIterator {

    private Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        this.stack = new Stack<>();

        while(root != null) {

            stack.push(root);
            root = root.left;
        }
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return this.stack.size() > 0;
    }

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

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