学步园

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.

题目题意感觉不好理解，查了下大概是按中序遍历把节点非递归输出

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    stack<TreeNode*>s;
    BSTIterator(TreeNode *root) {
        while(root != NULL){
            s.push(root);
            root = root->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !s.empty();
    }

    /** @return the next smallest number */
    int next() {
        TreeNode * tmp = s.top();
         s.pop();
         if(tmp->right != NULL){
            TreeNode* p = tmp->right;
            while(p){
                s.push(p);
                p = p->left;
            }
         }
         return tmp->val;
    }
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */