程序员面试100题--01

#include <iostream>
#include <string>
#include <memory.h>
#include <vector>
#include <sstream>
#include <math.h>
using namespace std;

/*
思路：
递归的转化左子树以及右子树，然后将当前节点作为左子树链表的末尾节点，
作为右子树的链表的开头结点.
但是得记住需要左子树的最后一个节点以及右子树的第一个节点
*/
struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x):val(x),left(NULL),right(NULL)
    {

    }
};
TreeNode* ConvertNode(TreeNode *pNode,bool isRight)
{
    if(pNode == NULL)
        return NULL;
    TreeNode* left = pNode->left;
    TreeNode* right = pNode->right;
    TreeNode* preturn = NULL;
    TreeNode* root = pNode;
    if(isRight)
    {
        while(root->left)
            root = root->left;
        preturn = root;
    }
    else
    {
        while(root->right)
            root = root->right;
        preturn = root;
    }

    TreeNode* pLeft = NULL;
    if(left != NULL)
    {
        pLeft = ConvertNode(left);
    }
    if(pLeft != NULL)
        pLeft->right = pNode;
    pNode->left = pLeft;

    TreeNode* pRight = NULL;
    if(right != NULL)
    {
        pRight = ConvertNode(right);
    }
    if(pRight != NULL)
        pRight->left = pNode;
    pNode->right = pRight;
    return preturn;
}

int main()
{

    return 0;
}

 题目描述：把二叉排序树转换成排序双向链表