LeetCode上的写一个二叉查找树的由小到大顺序的迭代器

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.

Note:

next()

and

hasNext()

should run in average O(1) time and uses O(h) memory, whereh is the height of the tree.

思路及实现：

 

#include<iostream>

#include<stack>

using namespace std;

struct TreeNode

{

    int val;

 TreeNode *left;

 TreeNode *right;

 TreeNode(int x) : val(x), left(NULL),right(NULL) {}

};

/*

BST是二叉查找数，左子数上的元素都小于根节点，右子数上的元素大于根节点

按照左子树，根，右子树的次序,就是一个从小到大的顺序

*/

class BSTIterator

{

public:

 BSTIterator(TreeNode *root)

 {

     TreeNode* pCurNode = root;

     while(pCurNode)

     {

            nodeStack.push(pCurNode);

            pCurNode = pCurNode->left;

     }

 }

 bool hasNext()

 {

        //栈空表示没有节点要访问了

        return !nodeStack.empty();

 }

 int next()

 {

     //每次进next()的时候，都保证当前节点是最左下,则其左子树已经没有节点了

        TreeNode* pCurNode = nodeStack.top();

        int value = pCurNode->val;

  

       //保存完右子树，弹出当前节点，

       pCurNode = pCurNode->right;

      nodeStack.pop();

        //如果右子树不为空，将其沿左下方的所有节点依次加入栈

        while(pCurNode != NULL)

        {

            nodeStack.push(pCurNode);

            pCurNode = pCurNode->left;

        }

  

       return value;

 }

private:

 stack<TreeNode*> nodeStack;

};

int main(int argc, char** argv)

{

    TreeNode* node6 = new TreeNode(6);

    TreeNode* node3 = new TreeNode(3);

    TreeNode* node9 = new TreeNode(9);

    TreeNode* node2 = new TreeNode(2);

    TreeNode* node5 = new TreeNode(5); 

    TreeNode* node7 = new TreeNode(7); 

    TreeNode* node11 = new TreeNode(11);

    TreeNode* node13 = new TreeNode(13);

    node6->left = node3;

   node6->right = node9;

   node3->left = node2;

   node3->right = node5;

   node9->left = node7;

   node9->right = node11;

   node11->right = node13;

   BSTIterator iter = BSTIterator(node6);

   while( iter.hasNext())

   {

        cout << iter.next() << endl;

   }

}