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C++中怎么利用LeetCode實現(xiàn)二叉搜索樹迭代器,很多新手對此不是很清楚,為了幫助大家解決這個難題,下面小編將為大家詳細(xì)講解,有這方面需求的人可以來學(xué)習(xí)下,希望你能有所收獲。
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.
這道題主要就是考二叉樹的中序遍歷的非遞歸形式,需要額外定義一個棧來輔助,二叉搜索樹的建樹規(guī)則就是左<根<右,用中序遍歷即可從小到大取出所有節(jié)點。代碼如下:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class BSTIterator { public: BSTIterator(TreeNode *root) { while (root) { s.push(root); root = root->left; } } /** @return whether we have a next smallest number */ bool hasNext() { return !s.empty(); } /** @return the next smallest number */ int next() { TreeNode *n = s.top(); s.pop(); int res = n->val; if (n->right) { n = n->right; while (n) { s.push(n); n = n->left; } } return res; } private: stack<TreeNode*> s; }; /** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */
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