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.
s思路： 1. 遍歷樹，用iterative。in-order遍歷的套路。
class BSTIterator {PRivate: stack<TreeNode*> ss; TreeNode* cur;public: BSTIterator(TreeNode *root) { cur=root; while(cur){ ss.push(cur); cur=cur->left; } } /** @return whether we have a next smallest number */ bool hasNext() { return !ss.empty(); } /** @return the next smallest number */ int next() { cur=ss.top(); ss.pop(); int res=cur->val; cur=cur->right; while(cur){ ss.push(cur); cur=cur->left; } return res; }};/** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */新聞熱點(diǎn)
疑難解答
圖片精選
網(wǎng)友關(guān)注
