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.
解题思路:
其实就是按照中序遍历的顺序出列,有两种实现思路:
一、构造BST的时候直接预处理下,构造出一条按照中序遍历的list,每次读取list的元素即可。
二、不进行预处理,按照中序遍历的思路使用Stack动态的进行处理
这里我们实现思路二:
public class BSTIterator { private Stack<TreeNode> stack=new Stack<TreeNode>(); public BSTIterator(TreeNode root) { if (root != null) pushLeft(root); } /** @return whether we have a next smallest number */ public boolean hasNext() { return !stack.isEmpty(); } /** @return the next smallest number */ public int next() { TreeNode top = stack.peek(); stack.pop(); if(top.right!=null) pushLeft(top.right); return top.val; } public void pushLeft(TreeNode root) { stack.push(root); TreeNode rootTemp = root.left; while (rootTemp != null) { stack.push(rootTemp); rootTemp = rootTemp.left; } } }