173.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.

Callingnext()will return the next smallest number in the BST.

Note:next()andhasNext()should run in average O(1) time and uses O(h) memory, wherehis the height of the tree.

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

public class BSTIterator {

    public BSTIterator(TreeNode root) {

    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {

    }

    /** @return the next smallest number */
    public int next() {

    }
}

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

inorder traversal, non-recursion,

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


public class BSTIterator {
    Stack<TreeNode> stack;
    TreeNode curt;
    public BSTIterator(TreeNode root) {
        stack = new Stack<>();
        curt = root;
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        if (curt != null || !stack.isEmpty()) {
            return true;
        }
        return false;
    }

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

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