Binary Search Tree Iterator

Design an iterator over a binary search tree with the following rules:

• Elements are visited in ascending order (i.e. an in-order traversal)
• next() and hasNext() queries run in O(1) time inaverage.  

Example

For the following binary search tree, in-order traversal by using iterator is[1, 6, 10, 11, 12]

10 /    \1      11 \       \  6       12

Challenge

Extra memory usage O(h), h is the height of the tree.

Super Star: Extra memory usage O(1)

 

SOLUTION：

这题啊，说实话，背诵题，首先背下来，再说理解。

开始说这个题，题本身来说，不难，就是一个inorder遍历，不过要用iterator来写（这里插一句，什么是iterator呢？ 迭代器基本有俩功能，next(),hasNext()，输出下一个元素，不过考虑到原始数据如果从list换成arraylist这种类似情况，而导致从新写一个for循环带来的麻烦，研究出一个迭代器，所有循环在迭代器内部完成），用iterator就是分开写这个这个程序，输出中序遍历的下一个元素。

思路就是中序遍历，具体看代码：

/** * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } * Example of iterate a tree: * BSTIterator iterator = new BSTIterator(root); * while (iterator.hasNext()) { *    TreeNode node = iterator.next(); *    do something for node * }  */public class BSTIterator {    //@param root: The root of binary tree.    private Stack
        
          stack = new Stack
         
          ();    private TreeNode current;    public BSTIterator(TreeNode root) {        current = root;    }    //@return: True if there has next node, or false    public boolean hasNext() {        return (current != null || !stack.isEmpty());    }        //@return: return next node    public TreeNode next() {        while (current != null){            stack.push(current);            current = current.left;        }        current = stack.pop();        TreeNode node = current;        current = current.right;        return node;    }}