二叉搜索树与双向链表 深度搜索 遍历二叉树 双向链表

https://leetcode-cn.com/problems/er-cha-sou-suo-shu-yu-shuang-xiang-lian-biao-lcof/

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""" # Definition for a Node. class Node(object):     def __init__(self, val, left=None, right=None):         self.val = val         self.left = left         self.right = right """     class Solution(object):     def treeToDoublyList(self, root):         """         :type root: Node         :rtype: Node         """         def preorder(current):             if current is None:                 return             preorder(current.left)             if self.pre is None:                 self.head = current             else:                 self.pre.right = current                 current.left = self.pre             self.pre = current             preorder(current.right)         if root is None:             return         self.pre = None         preorder(root)         self.head.left = self.pre         self.pre.right = self.head         return self.head

　　https://leetcode-cn.com/problems/er-cha-sou-suo-shu-yu-shuang-xiang-lian-biao-lcof/solution/mian-shi-ti-36-er-cha-sou-suo-shu-yu-shuang-xian-5/

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class Solution:     def treeToDoublyList(self, root: 'Node') -> 'Node':         def dfs(cur):             if not cur: return             dfs(cur.left) # 递归左子树             if self.pre: # 修改节点引用                 self.pre.right, cur.left = cur, self.pre             else: # 记录头节点                 self.head = cur             self.pre = cur # 保存 cur             dfs(cur.right) # 递归右子树                   if not root: return         self.pre = None         dfs(root)         self.head.left, self.pre.right = self.pre, self.head         return self.head   作者：jyd 链接：https://leetcode-cn.com/problems/er-cha-sou-suo-shu-yu-shuang-xiang-lian-biao-lcof/solution/mian-shi-ti-36-er-cha-sou-suo-shu-yu-shuang-xian-5/ 来源：力扣（LeetCode） 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。