题目如下：
解题思路：从两个根节点开始分别同步遍历两棵树，如果左右子树相同则表示不用交换；如果左右互相等于对方则交换；否则表示无法交换。
代码如下：
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None class Solution(object):
res = True
def verify(self,node1,node2):
leftV1 = None if node1.left == None else node1.left.val
leftV2 = None if node2.left == None else node2.left.val
rightV1 = None if node1.right == None else node1.right.val
rightV2 = None if node2.right == None else node2.right.val
if leftV1 == leftV2 and rightV1 == rightV2:
return 0
elif leftV1 == rightV2 and rightV1 == leftV2:
return 1
else:
return -1
def traverse(self,node1,node2):
if node1 == None or node2 == None:
return
ret = self.verify(node1,node2)
if ret == 0:
self.traverse(node1.left,node2.left)
self.traverse(node1.right, node2.right)
elif ret == 1:
node2.left,node2.right = node2.right,node2.left
self.traverse(node1.left, node2.left)
self.traverse(node1.right, node2.right)
else:
self.res = False def flipEquiv(self, root1, root2):
"""
:type root1: TreeNode
:type root2: TreeNode
:rtype: bool
"""
if (root1 == None) ^ (root2 == None):
return False
self.res = True
self.traverse(root1,root2)
return self.res