173. Binary Search Tree Iterator ---M

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.

 

Example:

BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false

 

Note:

• next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
• You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

A:

就是简单的用stack来记录当前的path.
思想跟iterative in-order traversal类似。   但这里，多了一个性质，就是该树是BST (但是题目里面没有用到)。

/**
 * 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:
    BSTIterator(TreeNode* root) {
        TreeNode* ptr = root;
        while(ptr){
            myStack.push(ptr);
            ptr= ptr->left;
        }
    }
    
    /** @return the next smallest number */
    int next() {
        auto res = myStack.top();
        myStack.pop();
        TreeNode * ptr = res->right;
        while(ptr){
            myStack.push(ptr);
            ptr= ptr->left;
        }
        return res->val;
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return not myStack.empty();
    }
private:
    stack<TreeNode*> myStack;
};

/**
 * 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();
 */

Mistakes:
1：  恩，错误是  致命的。

        TreeNode* ptr = root;
        while(ptr){
            myStack.push(ptr);
            ptr= ptr->left;
        }

   我写成了 

        TreeNode* ptr = root;
        while(ptr){
            myStack.push(ptr);

            if(ptr->left)
                ptr= ptr->left;
        }

造成了死循环