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QUESTION
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.
1ST TRY
前序遍历,记录下左子树的遍历路径
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ struct MyListNode { TreeNode* node; MyListNode *next; MyListNode(TreeNode* x) : node(x), next(NULL) {} }; class BSTIterator { public: BSTIterator(TreeNode *root) { currentRoot = root; leftTree = NULL; buildLeftTree(root); } /** @return whether we have a next smallest number */ bool hasNext() { if(leftTree) return true; else return false; } /** @return the next smallest number */ int next() { ret = leftTree->node->val; tmpListNode = leftTree; leftTree = leftTree->next; buildLeftTree(tmpListNode->node->right); delete[] tmpListNode; return ret; } void buildLeftTree(TreeNode *root) { if(!root) return; tmpTreeNode = root; while(tmpTreeNode) { MyListNode *newListNode = new MyListNode(tmpTreeNode); newListNode->next = leftTree; leftTree = newListNode; tmpTreeNode = tmpTreeNode->left; } } private: TreeNode *tmpTreeNode; TreeNode *currentRoot; MyListNode *tmpListNode; MyListNode *leftTree; int ret; }; /** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */
Result: Accepted
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原文地址:http://www.cnblogs.com/qionglouyuyu/p/4212147.html