leetcode之Binary Search Tree Iterator

题目：

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.

解答：

由于BST的中序遍历就是从小到大的顺序， 所以直接进行中序遍历即可，为了满足复杂度要求，需要将递归改为迭代过程，用栈实现

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