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.
基本思路:
事实上就是中序遍历的非递实现。
这next smallest意思就是。从小到大逐个返回。 第一次看到,居然理解反了。还以为是从大到小逐个返回。
/** * 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() { auto root = s_.top(); s_.pop(); auto ans = root->val; root = root->right; while (root) { s_.push(root); root = root->left; } return ans; } private: stack<TreeNode *> s_; }; /** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */