题解 | #二叉搜索树与双向链表#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

#
# 
# @param pRootOfTree TreeNode类 
# @return TreeNode类
#
class Solution:
    def Convert(self , pRootOfTree ):
        
        def inorder(root,order):
            if root ==None:
                return None
            inorder(root.left, order)
            order.append(root)   #[list] order里加入的是一个节点而非value
            inorder(root.right,order)
            
        
        if not pRootOfTree:
            return None

        list_order=[]
        inorder(pRootOfTree, list_order)
        if len(list_order)==1:
            return pRootOfTree
        for i in range(len(list_order)):
            if i==0:
                list_order[i].left=None
            else:
                list_order[i].left=list_order[i-1]
            if i==len(list_order)-1:
                list_order[i].right=None
            else:
                list_order[i].right=list_order[i+1]
        return list_order[0]
        # write code here