[Java]leetcode173 Binary Search Tree Iterator

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：写一个二叉查找树的迭代器，实现hasNext()和next()的功能。next()每次返回二叉树中未访问的最小值。要求的平均时间复杂度是O(1)和空间复杂度是O(h)。

解题思路：next()每次返回二叉树中未访问的最小值。也就是将二叉树中序遍历，并将对应值输出。这里面注意的是空间和时间的复杂度。

public class BSTIterator {//将中序遍历的功能嵌查在整个程序中
    TreeNode current;
    Stack<TreeNode> stack;
    public BSTIterator(TreeNode root) {
        current=root;
        stack=new Stack<TreeNode>();
        while(current!=null)//因为只可能是左节点才是最小值，将所有左子树节点入栈，保证空间复杂度是O(h)
        {
            stack.push(current);
            current=current.left;
        }
    }
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
     return !stack.isEmpty();
    }
    /** @return the next smallest number */
    public int next() {
      current=stack.pop();
      int res=current.val;//这一步已经得到最小值
      current=current.right;//但要考虑到右子树的遍历
      while(current!=null)
       {
         stack.push(current);
         current=current.left;
       }
       return res;
      }
}