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# Binary Search Tree Iterator LeetCode Solution

## Problem – Binary Search Tree Iterator

Implement the `BSTIterator` class that represents an iterator over the in-order traversal of a binary search tree (BST):

• `BSTIterator(TreeNode root)` Initializes an object of the `BSTIterator` class. The `root` of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
• `boolean hasNext()` Returns `true` if there exists a number in the traversal to the right of the pointer, otherwise returns `false`.
• `int next()` Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to `next()` will return the smallest element in the BST.

You may assume that `next()` calls will always be valid. That is, there will be at least a next number in the in-order traversal when `next()` is called.

Example 1:

``````Input
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output
[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next();    // return 3
bSTIterator.next();    // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 20
bSTIterator.hasNext(); // return False
``````

Constraints:

• The number of nodes in the tree is in the range `[1, 105]`.
• `0 <= Node.val <= 106`
• At most `105` calls will be made to `hasNext`, and `next`.

• Could you implement `next()` and `hasNext()` to run in average `O(1)` time and use `O(h)` memory, where `h` is the height of the tree?

### Binary Search Tree Iterator LeetCode Solution in Java

``````public class BSTIterator {
private Stack<TreeNode> stack = new Stack<TreeNode>();

public BSTIterator(TreeNode root) {
pushAll(root);
}

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

/** @return the next smallest number */
public int next() {
TreeNode tmpNode = stack.pop();
pushAll(tmpNode.right);
return tmpNode.val;
}

private void pushAll(TreeNode node) {
for (; node != null; stack.push(node), node = node.left);
}
}
``````

### Binary Search Tree Iterator LeetCode Solution in C++

``````class BSTIterator {
stack<TreeNode *> myStack;
public:
BSTIterator(TreeNode *root) {
pushAll(root);
}

/** @return whether we have a next smallest number */
bool hasNext() {
return !myStack.empty();
}

/** @return the next smallest number */
int next() {
TreeNode *tmpNode = myStack.top();
myStack.pop();
pushAll(tmpNode->right);
return tmpNode->val;
}

private:
void pushAll(TreeNode *node) {
for (; node != NULL; myStack.push(node), node = node->left);
}
};
``````

### Binary Search Tree Iterator LeetCode Solution in Python

``````class BSTIterator:
# @param root, a binary search tree's root node
def __init__(self, root):
self.stack = list()
self.pushAll(root)

# @return a boolean, whether we have a next smallest number
def hasNext(self):
return self.stack

# @return an integer, the next smallest number
def next(self):
tmpNode = self.stack.pop()
self.pushAll(tmpNode.right)
return tmpNode.val

def pushAll(self, node):
while node is not None:
self.stack.append(node)
node = node.left
``````
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