# Why BST and not Binary Tree?

• I solved this problem by using a stack (other users have already posted similar solutions).

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

public class BSTIterator {

Deque<TreeNode> stack = new ArrayDeque<>();

public BSTIterator(TreeNode root) {
while (root != null) {
root = root.left;
}
}

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

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

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

However, in this solution, I don't take into account that I am traversing a binary search tree; the solution should work with any binary tree.

Therefore, given that the problem statement mentions BST, I am wondering whether there is any way to take advantage of the BST properties and come up with a more efficient solution.

• The next() function is supposed to return the next smallest number. A BST makes that very easy to do since an in order traversal of the tree works. It won't be so easy with just a binary tree since there's no ordering or structure to the tree.

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