题解 | #二叉搜索树与双向链表#

class Solution:
    def Convert(self , pRootOfTree ):
        # write code here
        if not pRootOfTree:
            return 
        self.result = []
        self.midOrder(pRootOfTree)
        
        for i in range(1, len(self.result)):
            self.result[i].left = self.result[i-1]
            self.result[i-1].right = self.result[i]
        return self.result[0]
    
    def midOrder(self, p): # 无论是先序、中序还是后序，递归出口都在头部
        if not p: return
        
        self.midOrder(p.left)
        self.result.append(p)
        self.midOrder(p.right)