Binary Search Tree Iterator

LeetCode 173–Binary Search Tree Iterator

Problem

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.

题目大意

实现一个用二叉排序树根节点初始化的二叉排序树迭代器。
调用 next() 方法返回二叉排序树中最小的下一个数。
注意：next() 和 hasNext()方法的平均时间复杂度为O(1) ，空间复杂度为 O(h)，h代表树的高度。

思路

该题目考察二叉树的中序遍历，步骤如下：    1. 初始化时，根节点进栈，左子树进栈，初始化结束。     2. 调用 hasNext()方法时，判断栈是否为空，若为空，则返回false，否则，返回true。    3. 调用next()方法时，栈顶元素出栈。并判断出栈元素是否有又子树，若存在，则右子树进栈，右子树的左子树进栈。返回出栈元素的值。

代码实现–Java

/** * 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 = new Stack<TreeNode>();    public BSTIterator(TreeNode root) {        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 root = stack.pop();        TreeNode node = root.right;        while(node != null){            stack.push(node);            node = node.left;        }        return root.val;    }}/** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */

结果