[LeetCode] Binary Search Tree Iterator，leetcodeiterator

[LeetCode] Binary Search Tree Iterator，leetcodeiterator

Binary Search Tree Iterator

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.

 

解題思路：

運用棧的技術，將當前未被訪問的最左邊的一條路徑入棧，每次取值的時候將棧頂元素彈出，並將棧頂元素的右子樹的最左邊一條路徑入棧。hasNext()只需要看棧是否為空白即可。但是有個問題，似乎next()函數的時間複雜度並不是O(1)，因為我們要維護棧，因此是O(nlgn)。不知有什麼更為好的方法沒有。我還犯了一個小錯誤，就是stack<TreeNode*> stack。不能取名叫stack呀。

/** * 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) {        pushLeftChildIntoStack(root);    }    /** @return whether we have a next smallest number */    bool hasNext() {        return !s.empty();    }    /** @return the next smallest number */    int next() {        TreeNode* node=s.top();        s.pop();        pushLeftChildIntoStack(node->right);        return node->val;    }private:    stack<TreeNode*> s;    void pushLeftChildIntoStack(TreeNode* node){        while(node!=NULL){            s.push(node);            node=node->left;        }    }};/** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */