Binary Search Tree Iterator(lintcode 86)

Description

Design an iterator over a binary search tree with the following rules:

• Elements are visited in ascending order (i.e. an in-order traversal)
• next() and hasNext() queries run in O(1) time in average.

Example

For the following binary search tree, in-order traversal by using iterator is [1, 6, 10, 11, 12]

   10
 /    \
1      11
 \       \
  6       12

Interface

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Example of iterate a tree:
 * BSTIterator iterator = new BSTIterator(root);
 * while (iterator.hasNext()) {
 *    TreeNode node = iterator.next();
 *    do something for node
 * } 
 */
public class BSTIterator {
    //@param root: The root of binary tree.
    public BSTIterator(TreeNode root) {
        // write your code here
    }

    //@return: True if there has next node, or false
    public boolean hasNext() {
        // write your code here
    }

    //@return: return next node
    public TreeNode next() {
        // write your code here
    }
}

Solution

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Example of iterate a tree:
 * BSTIterator iterator = new BSTIterator(root);
 * while (iterator.hasNext()) {
 *    TreeNode node = iterator.next();
 *    do something for node
 * } 
 */
public class BSTIterator {
    //@param root: The root of binary tree.
    private Stack<TreeNode> stack = new Stack<>();
    private TreeNode curr;

    public BSTIterator(TreeNode root) {
        // write your code here
        curr = root;
    }

    //@return: True if there has next node, or false
    public boolean hasNext() {
        // write your code here
        return !(curr == null && stack.isEmpty());
    }

    //@return: return next node
    public TreeNode next() {
        // write your code here
        while (curr != null) {
            stack.push(curr);
            curr = curr.left;
        }

        curr = stack.pop();
        TreeNode result = curr;
        curr = curr.right;

        return result;
    }
}