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.

Solution

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

public class BSTIterator {

    List<TreeNode> lst;
    int cur;

    public BSTIterator(TreeNode root) {
        lst = new ArrayList<>();
        travse(root, lst);

        cur = -1;
    }


    private void travse(TreeNode root, List<TreeNode> lst) {
        if (root == null) {
            return;
        }

        travse(root.left, lst);
        lst.add(root);
        travse(root.right, lst);
    }

    /**
     * @return whether we have a next smallest number
     */
    public boolean hasNext() {
        return cur + 1 < lst.size();
    }

    /**
     * @return the next smallest number
     */
    public int next() {
        if (hasNext()) {
            cur ++;
            return lst.get(cur).val;
        }

        return 0;

    }
}

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