Binary Search Tree Iterator问题及解法

问题描述：

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.

问题分析：

由于BST遍历是根的中序遍历，所以我们这里使用一种数据结构stack，先把root及其之后左节点的left依次入栈，每次判断是否有下一个时只需要判断stack是否为空即可，而next()操作则需要取出栈顶元素root，后将root->right及其之后的左节点的left入栈，返回root->val即可。

过程详见代码：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class BSTIterator {private:    stack<TreeNode*> st;public:    BSTIterator(TreeNode *root) {        find_left(root);    }    /** @return whether we have a next smallest number */    bool hasNext() {        if (st.empty())            return false;        return true;    }    /** @return the next smallest number */    int next() {        TreeNode* top = st.top();        st.pop();        if (top->right != NULL)            find_left(top->right);                    return top->val;    }        /** put all the left child() of root */    void find_left(TreeNode* root)    {        TreeNode* p = root;        while (p != NULL)        {            st.push(p);            p = p->left;        }    }};/** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */