剑指offer---二叉搜索树与双向链表

题目：二叉搜索树与双向链表

要求：输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。

/*
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)
    {
        
    }
};

解题代码：

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree) {
        // pLastNodeInList 指向双向链表的最后一个节点
        TreeNode* pLastNodeInList = nullptr;
        ConvertNode(pRootOfTree, pLastNodeInList);

        // 返回双向链表的头结点
        TreeNode* pHeadOfList = pLastNodeInList;
        while(pHeadOfList != nullptr && pHeadOfList->left != nullptr)
            pHeadOfList = pHeadOfList->left;
        return pHeadOfList;
    }

private:
    void ConvertNode(TreeNode* pNode, TreeNode * &pLastNodeInList){
        if(pNode == nullptr)
            return ;

        TreeNode* pCurrent = pNode;
        if(pCurrent->left != nullptr)
            ConvertNode(pCurrent->left, pLastNodeInList);

        pCurrent->left = pLastNodeInList;
        // 若此时链表含有节点，则将链表最后一个节点的右指针指向pCurrent节点
        if(pLastNodeInList != nullptr)
            pLastNodeInList->right = pCurrent;
        // 更新指向链表最后一个节点的指针
        pLastNodeInList = pCurrent;

        if(pCurrent->right != nullptr)
            ConvertNode(pCurrent->right, pLastNodeInList);
    }
};