# Question

Formatted question description: https://leetcode.ca/all/173.html

 173	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.

@tag-tree


# Algorithm

The non-recursive form of the in-order traversal of the binary tree requires an additional definition of a stack to assist. The building rule of the binary search tree is left<root<right, and the in-order traversal can extract all nodes from small to large.

# Code

Java

• import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Stack;

public class Binary_Search_Tree_Iterator {

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

Stack<TreeNode> sk;

// @note:@memorize: important feature, min is always going left branch to leaf
public BSTIterator(TreeNode root) {
sk = new Stack<>();

// all the way to leftmost leaf
while (root != null) {
sk.push(root);
root = root.left;
}

}

/** @return whether we have a next smallest number */
public boolean hasNext() {
return !sk.isEmpty();
}

/** @return the next smallest number */
public int next() {
TreeNode minNode = sk.pop();

TreeNode current = minNode;
// update stack, possible next time min is frmo its right-then-left branch
if (current.right != null) {
current = current.right;

// same logic as in constructor
while (current != null) {
sk.push(current.left);
current = current.left;
}
}

return minNode.val;
}
}

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

• // OJ: https://leetcode.com/problems/binary-search-tree-iterator/
// Time: O(1) amortized
// Space: O(H)
class BSTIterator {
private:
stack<TreeNode*> s;
void pushNodes(TreeNode *node) {
while (node) {
s.push(node);
node = node->left;
}
}
public:
BSTIterator(TreeNode* root) {
pushNodes(root);
}
int next() {
auto node = s.top();
s.pop();
pushNodes(node->right);
return node->val;
}
bool hasNext() {
return s.size();
}
};

• # 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.p = None
self.stack = []
if root:
self.stack.append((1, root))

def hasNext(self):
"""
:rtype: bool
"""
return len(self.stack) > 0

def next(self):
"""
:rtype: int
"""
stack = self.stack
while stack:
p = stack.pop()
if not p:
continue
if p == 0:
return p.val
else:
l = []
if p.right:
l.append((1, p.right))
l.append((0, p))
if p.left:
l.append((1, p.left))
stack.extend(l)

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