标签:views 调整 solution 更新 amp interview tree 二叉搜索树 int
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:
1 class Solution { 2 public: 3 TreeNode* Convert(TreeNode* pRootOfTree) { 4 // pLastNodeInList 指向双向链表的最后一个节点 5 TreeNode* pLastNodeInList = nullptr; 6 ConvertNode(pRootOfTree, pLastNodeInList); 7 8 // 返回双向链表的头结点 9 TreeNode* pHeadOfList = pLastNodeInList; 10 while(pHeadOfList != nullptr && pHeadOfList->left != nullptr) 11 pHeadOfList = pHeadOfList->left; 12 return pHeadOfList; 13 } 14 15 private: 16 void ConvertNode(TreeNode* pNode, TreeNode * &pLastNodeInList){ 17 if(pNode == nullptr) 18 return ; 19 20 TreeNode* pCurrent = pNode; 21 if(pCurrent->left != nullptr) 22 ConvertNode(pCurrent->left, pLastNodeInList); 23 24 pCurrent->left = pLastNodeInList; 25 // 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 26 if(pLastNodeInList != nullptr) 27 pLastNodeInList->right = pCurrent; 28 // 更新指向链表最后一个节点的指针 29 pLastNodeInList = pCurrent; 30 31 if(pCurrent->right != nullptr) 32 ConvertNode(pCurrent->right, pLastNodeInList); 33 } 34 };
标签:views 调整 solution 更新 amp interview tree 二叉搜索树 int
原文地址:https://www.cnblogs.com/iwangzhengchao/p/9875766.html