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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.
思路：
说白了，就是把中序遍历拆成几个部分写。
我的代码：
class BSTIterator { public: BSTIterator(TreeNode *root) { pCur = root; while(pCur != NULL) { v.push_back(pCur); pCur = pCur->left; } } /** @return whether we have a next smallest number */ bool hasNext() { return (!v.empty() || NULL != pCur); } /** @return the next smallest number */ int next() { int num; TreeNode * tmp = v.back(); v.pop_back(); num = tmp->val; pCur = tmp->right; while(pCur != NULL) { v.push_back(pCur); pCur = pCur->left; } return num; } private: vector<TreeNode *> v; TreeNode * pCur; };
大神更精简的代码： 经验，把相同功能的代码放在一起可以简化代码。
public class BSTIterator { Stack<TreeNode> stack = null ; TreeNode current = null ; public BSTIterator(TreeNode root) { current = root; stack = new Stack<> (); } /** @return whether we have a next smallest number */ public boolean hasNext() { return !stack.isEmpty() || current != null; } /** @return the next smallest number */ public int next() { while (current != null) { stack.push(current); current = current.left ; } TreeNode t = stack.pop() ; current = t.right ; return t.val ; } }
【leetcode】Binary Search Tree Iterator（middle）
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原文地址：http://www.cnblogs.com/dplearning/p/4475866.html
