剑指offer 二叉搜索树与双向链表

题目描述

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)
    {
        TreeNode *pLastNodeInTree = NULL;
        TreeConvert(pRootOfTree, pLastNodeInTree);
        TreeNode *pFirstNodeInTree = pLastNodeInTree;
        while (pFirstNodeInTree != NULL && pFirstNodeInTree->left != NULL) {
            pFirstNodeInTree = pFirstNodeInTree->left;
        }
        return pFirstNodeInTree;
    }
    void TreeConvert(TreeNode* pNode, TreeNode* &pLastNodeInTree) {
        if (pNode == NULL) {
            return;
        }
        TreeNode* pCurrent = pNode;
        if (pNode->left != NULL) {
            TreeConvert(pNode->left, pLastNodeInTree);
        }
        pCurrent->left = pLastNodeInTree;
        if (pLastNodeInTree != NULL)
            pLastNodeInTree->right = pCurrent;
        pLastNodeInTree = pCurrent;
        if (pNode->right != NULL) {
            TreeConvert(pNode->right, pLastNodeInTree);
        }
    }
};