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 mid_DFS(self , pRootOfTree):
global output
if not pRootOfTree:
return
self.mid_DFS(pRootOfTree.left)
output.right=pRootOfTree
pRootOfTree.left=output
output=pRootOfTree
self.mid_DFS(pRootOfTree.right)
def Convert(self , pRootOfTree ):
# write code here
if not pRootOfTree:
return
global output
output=TreeNode(None)
self.mid_DFS(pRootOfTree)
while output.left.val!=None:
output=output.left
output.left=None
return output
