二叉搜索树与双向链表

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。

递归版: 树的先中后序的处理就理解成按固定次序处理一个序列, 从大到小, 即中序, 通过指针传回上次递归的值(递归还能这么玩)

class Solution {
public:
    void convertNode(TreeNode* pNode, TreeNode **pLastInList) {
        if (nullptr == pNode)    return;

		TreeNode *pCurrent = pNode;		// 可以使用pNode来代替pCurrent
        
        if (nullptr != pCurrent->left) {
            convertNode(pCurrent->left, pLastInList);
        }
        
        pCurrent->left = *pLastInList;
        if (nullptr != *pLastInList) {
            (*pLastInList)->right = pCurrent;
        }
        *pLastInList = pCurrent;
        
        if (nullptr != pCurrent->right) {
            convertNode(pCurrent->right, pLastInList);
        }
    }
    
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        TreeNode *pLastInList = nullptr;
        
        convertNode(pRootOfTree, &pLastInList);
        
        TreeNode *pHead = pLastInList;
        while ((nullptr != pHead) && (nullptr != pHead->left)) {
            pHead = pHead->left;
        }
        
        return pHead;
    }
};

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/