[LeetCode] 173. Binary Search Tree Iterator Java

题目：

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.

题意及分析：给出一个二叉排序树，求一个该二叉树的遍历器，满足：(1)hasNext()判断树中是否存在一个除了当前数的最小数。 (2)next()返回树中除了当前数的最小数。(3)要求o(1)的时间复杂度和o(h)的空间复杂度。使用一个栈保存即可，初始保存从根节点到最左节点的值，对于next，如果stack非空就存在hasnext smallest number；对于next，下一个最小的就是栈顶的值（因为该栈顶点的左子节点就是当前点，而栈顶点的右子树上的点肯定比栈顶点大），取出栈顶的点，然后对该点的右子节点左上面的操作(即遍历该点到该点的最左叶节点)。

代码：

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

public class BSTIterator {

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

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

    /** @return the next smallest number */
    public int next() {
    	TreeNode node = stack.pop();
        TreeNode cur = node;
        // traversal right branch
        if(cur.right != null){
            cur = cur.right;
            while(cur != null){
                stack.push(cur);
                if(cur.left != null)
                    cur = cur.left;
                else
                    break;
            }
        }
        return node.val;
    }
}

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