acwing 49. 二叉搜索树与双向链表
地址:https://www.acwing.com/problem/content/87/
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。
注意:
- 需要返回双向链表最左侧的节点。
例如,输入下图中左边的二叉搜索树,则输出右边的排序双向链表。
解法
树的处理 一半都是递归 分为 根 树的左子树 和树的右子树
子树也是一棵树 进行递归处理 向上返回一个双链表 返回链表的头尾
最后全部转化链表
代码
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* rethead = NULL; TreeNode* gleft = NULL; TreeNode* gright = NULL; void convertInner(TreeNode* root) { if (NULL == root) return; if (root->val < rethead->val) rethead = root; if (root->left == NULL && root->right == NULL) { gleft = root; gright = root; return; } else if (root->left != NULL && root->right == NULL) { convertInner(root->left); gright->right = root; root->left = gright; gright = root; } else if (root->right != NULL && root->left == NULL) { convertInner(root->right); gleft->left = root; root->right = gleft; gleft = root; } else if (root->right != NULL && root->left != NULL) { convertInner(root->left); gright->right = root; root->left = gright; TreeNode* leftcopy = gleft; convertInner(root->right); gleft->left = root; root->right = gleft; gleft = leftcopy; } } TreeNode* convert(TreeNode* root) { if (NULL == root) return NULL; rethead = root; if (root->left == NULL && root->right == NULL) return root; if (root->left != NULL && root->right == NULL) { convertInner(root->left); gright->right = root; root->left = gright; } else if (root->right != NULL && root->left == NULL) { convertInner(root->right); gleft->left = root; root->right = gleft; } else if (root->right != NULL && root->left != NULL) { convertInner(root->left); gright->right = root; root->left = gright; convertInner(root->right); gleft->left = root; root->right = gleft; } return rethead; } };
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* rethead = NULL; TreeNode* gleft = NULL; TreeNode* gright = NULL; void convertInner(TreeNode* root) { if (NULL == root) return; if (root->val < rethead->val) rethead = root; if (root->left == NULL && root->right == NULL) { gleft = root; gright = root; return; } else if (root->left != NULL && root->right == NULL) { convertInner(root->left); gright->right = root; root->left = gright; gright = root; } else if (root->right != NULL && root->left == NULL) { convertInner(root->right); gleft->left = root; root->right = gleft; gleft = root; } else if (root->right != NULL && root->left != NULL) { convertInner(root->left); gright->right = root; root->left = gright; TreeNode* leftcopy = gleft; convertInner(root->right); gleft->left = root; root->right = gleft; gleft = leftcopy; } } TreeNode* convert(TreeNode* root) { if (NULL == root) return NULL; rethead = root; if (root->left == NULL && root->right == NULL) return root; if (root->left != NULL && root->right == NULL) { convertInner(root->left); gright->right = root; root->left = gright; } else if (root->right != NULL && root->left == NULL) { convertInner(root->right); gleft->left = root; root->right = gleft; } else if (root->right != NULL && root->left != NULL) { convertInner(root->left); gright->right = root; root->left = gright; convertInner(root->right); gleft->left = root; root->right = gleft; } return rethead; } };
作 者: itdef
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B站算法视频题解
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qq 151435887
gitee https://gitee.com/def/
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欢迎转帖 请保持文本完整并注明出处
技术博客 http://www.cnblogs.com/itdef/
B站算法视频题解
https://space.bilibili.com/18508846
qq 151435887
gitee https://gitee.com/def/
欢迎c c++ 算法爱好者 windows驱动爱好者 服务器程序员沟通交流
如果觉得不错,欢迎点赞,你的鼓励就是我的动力