C++中怎么利用LeetCode实现二叉搜索树迭代器,很多新手对此不是很清楚,为了帮助大家解决这个难题,下面小编将为大家详细讲解,有这方面需求的人可以来学习下,希望你能有所收获。
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.
这道题主要就是考二叉树的中序遍历的非递归形式,需要额外定义一个栈来辅助,二叉搜索树的建树规则就是左<根<右,用中序遍历即可从小到大取出所有节点。代码如下:
/** * 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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