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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.
利用二叉树中序非递归遍历的思想,使用一个栈即可,代码如下:
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class BSTIterator { 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 p = stack.pop(); int re = p.val; p = p.right; if(p!=null) { stack.push(p); while(p.left!=null) { stack.push(p.left); p = p.left; } } return re; } } /** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */
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原文地址:http://www.cnblogs.com/mrpod2g/p/4329320.html