算法1

思路：非递归中序遍历，每次把根节点的左链放入栈中，如果左子树遍历完，将右子树的左链放到栈里

C++ 代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class BSTIterator {
public:
    stack<TreeNode*> stk;
    BSTIterator(TreeNode* root) {
        while(root)
        {
            stk.push(root);//每次把根节点放进来，根节点放进根节点的左链
            root=root->left;

        }
    }

    int next() {
        auto root=stk.top();//根节点是栈顶元素，将栈顶元素删掉
        stk.pop();
        int val=root->val;
        root=root->right;
        while(root)//根节点存在，把根节点的右子树左链存到栈里
        {
            stk.push(root);
            root=root->left;

        }
        return val;
    }

    bool hasNext() {
        return stk.size();
    }
};

/**
 * 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();
 */