二叉搜索树与双向链表 深度搜索 遍历二叉树 双向链表

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

"""
# 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/

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。

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