Leetcode 173. 二叉搜索树迭代器 (等价于中序非递归遍历二叉树）

如果直接用递归中序遍历存到数组中，然后输出，这样的时间复杂度是O(n)

为了只用O(h)的空间，我们采用非递归迭代策略。

思路：

能往左走就往左走，边走边入栈，直到不能走，弹出栈里元素往右走，重复之前操作。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
    
public:
    stack<TreeNode*> st;
    TreeNode* cur;

    BSTIterator(TreeNode* root) {
        cur = root;
    }

    
    /** @return the next smallest number */
    int next() {
        while(cur != NULL){
            st.push(cur);
            cur = cur -> left;
        }
        cur = st.top();
        st.pop();
        int val = cur->val;
        cur = cur->right;
        return val;
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return cur != NULL || !st.empty();
    }
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator* obj = new BSTIterator(root);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */