Description
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.
Solution
Inorder Traversal using Stack, time O(n), space O(h)
/**
* 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;
private TreeNode curr;
public BSTIterator(TreeNode root) {
stack = new Stack<>();
curr = root;
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return curr != null || !stack.empty();
}
/** @return the next smallest number */
public int next() {
while (curr != null) {
stack.push(curr);
curr = curr.left;
}
curr = stack.pop();
int val = curr.val;
curr = curr.right;
return val;
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
