[LeetCode 173] Binary Search Tree Iterator

Question

link

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.

            <div id="tags" class="btn btn-xs btn-warning">Show Tags</div>
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              <a class="btn btn-xs btn-primary" href="/tag/tree/">Tree</a>
              
              <a class="btn btn-xs btn-primary" href="/tag/stack/">Stack</a>
              
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Analysis

This is an extremely important question, if you are going for an interview. I repeat: this is an extremely important question, if you are going for an interview. If you do not remember it by heart, I will repeat again.

The solution of the iterator applies to a lot of related questions. So make sure you practise this question until you are perfect. You WILL BE ASKED this question at times.

You could read my other post [[Question] Iterator of Binary Search Tree].

Solution

We only need to keep 1 variable in RAM, that is a stack.

Code

public class BSTIterator {

    Stack<TreeNode> stack = new Stack<TreeNode>();

    public BSTIterator(TreeNode root) {
        while (root != null) {
            stack.push(root);
            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() {
        if (!hasNext()) {
            return 0;
        }
        TreeNode next = stack.pop();
        TreeNode node = next.right;
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
        return next.val;
    }
}