Binary Search Tree Iterator

描述

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.

分析

考察非递归的中序遍历。这道题本质上是写一个二叉树的中序遍历的迭代器。内部设置一个栈，初始化的时候，存储从根节点到最左叶子节点的路径。在遍历的过程中，每次从栈中弹出一个元素，作为当前的返回结果，同时探测一下当前节点是否存在右孩子，如果有，则进入右孩子，并把从该右孩子到最左叶子节点的所有节点入栈。

代码

// Binary Search Tree Iterator
public class BSTIterator {
    public BSTIterator(TreeNode root) {
        stack = new Stack<>();
        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() {
        final TreeNode node = stack.pop();
        if (node.right != null) {
            TreeNode p = node.right;
            while (p != null) {
                stack.push(p);
                p = p.left;
            }

        }
        return node.val;
    }
    private Stack<TreeNode> stack;
}