剑指offer 27：二叉搜索树与双向链表

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        //指向根节点左子树的最后一个节点
        TreeNode *pLastNodeInList=NULL;
        ConvertNode(pRootOfTree ,&pLastNodeInList);
        TreeNode * pHeadFirst=pLastNodeInList;
        //记录根节点的值
        while(pHeadFirst!=NULL && pHeadFirst->left!=NULL){
            pHeadFirst=pHeadFirst->left;
        }
        return pHeadFirst;
    }
    void ConvertNode(TreeNode* pRootOfTree ,TreeNode** pLastNodeInlist){
        if (pRootOfTree ==NULL){
            return ;
        }
        TreeNode * cur=pRootOfTree;
        if (cur->left!=NULL){
            ConvertNode(cur->left,pLastNodeInlist);
        }
        cur->left=*pLastNodeInlist;
        if(*pLastNodeInlist!=NULL){
            (*pLastNodeInlist)->right=cur;
        }
        (*pLastNodeInlist)=cur;
        if (cur->right!=NULL){
            ConvertNode(cur->right,pLastNodeInlist);
        }
    }
};