【leetcode】173. 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.

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思路：
用cur指向下一个节点。用my_stack（栈）只保存左子树的节点，目的是为了cur指针能够回退（因为是中序遍历所以需要回退）。如果下一个节点不存在，则cur会指向nullptr；否则将指向下一个节点。

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代码实现：

/**
 * 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*> my_stack;
    TreeNode *cur;
    BSTIterator(TreeNode* root) {
        if (root == nullptr){
            cur = nullptr;
        }else{
            while (root->left){
                my_stack.push(root);
                root = root->left;
            }
            cur = root;
        }
    }
    
    /** @return the next smallest number */
    int next() {
        int ret = cur->val;
        if (cur->right){
            cur = cur->right;
            while (cur->left){
                my_stack.push(cur);
                cur = cur->left;
            }
        }else{
            if (my_stack.empty()){
                cur = nullptr;
            }else{
                cur = my_stack.top();
                my_stack.pop();
            }
        }
        
        return ret;
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return cur != nullptr;
    }
};

/**
 * 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();
 */

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参考：
https://www.cnblogs.com/lightwindy/p/8607148.html