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题目:
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.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
代码:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class BSTIterator { private: vector<TreeNode*> list; int index; public: BSTIterator(TreeNode *root) { BSTIterator::treeToList(root, this->list); this->index = 0; } static void treeToList(TreeNode* root, vector<TreeNode*>& ret) { if ( !root ) return; BSTIterator::treeToList(root->left, ret); ret.push_back(root); BSTIterator::treeToList(root->right, ret); } /** @return whether we have a next smallest number */ bool hasNext() { return index < this->list.size(); } /** @return the next smallest number */ int next() { return list[index++]->val; } }; /** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */
tips:
这个题目的核心就是在于中序遍历 把BST转化成vector,这样可以实现题目的要求。
【Binary Search Tree Iterator 】cpp
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原文地址:http://www.cnblogs.com/xbf9xbf/p/4646397.html