LeetCode——Binary Search Tree Iterator

Description：

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.

实现一个迭代器来遍历二叉查找树。当然首先想到的方法就是中序遍历把有序元素保存在容器中，顺序操作。但是要求uses O(h) memory就不能这样干了。

这里用一个栈来做暂存器，先把所有左节点压栈，这时栈定就一定是最小的元素。之后出栈返回当前的栈顶元素，对于栈顶元素，把以其为顶点的右子树的所有左节点都压栈。以此类推，直到所有元素都遍历一遍。在uses O(h) memory的要求下完成了二叉查找树的遍历。

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

public class BSTIterator {
    public Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack = new Stack<TreeNode>();
        while(root != null) {
            stack.push(root);
            root = root.left;
        }

    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.empty();
    }

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

        TreeNode node = stack.peek();
        int nextVal = node.val;
        stack.pop();
        TreeNode t = node.right;
        while(t != null) {
            stack.push(t);
            t = t.left;
        }

        return nextVal;
    }
}

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