题解 | #二叉搜索树与双向链表#
二叉搜索树与双向链表
http://www.nowcoder.com/practice/947f6eb80d944a84850b0538bf0ec3a5
```# 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 ):
res=[]
if pRootOfTree is None:
return None
def zhongxu(root):
if root is None:
return
zhongxu(root.left)
res.append(root)
zhongxu(root.right)
zhongxu(pRootOfTree)
for i in range(len(res)-1):
res[i+1].left=res[i]
res[i].right=res[i+1]
return res[0]
##### 这题的思路就是先中序遍历一下,可以到中序节点的数组,在循环一下得到双向链表!