给定一个二叉树，检查它是否是镜像对称的。(随便看看吧,这个题真滴难-只有最后两个是对的前面的都不是对的)
# Definition for a binary tree node.
class TreeNode(object):
def __init__(self, val=None, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def isSymmetric1(self, root):
"""这个方法适合满二叉树,获取是空节点补TreeNode()的树
:type root: TreeNode
:rtype: bool
"""
temp = root
stack = list()
temp_list = [] # 几率元素个数
while temp or stack:
while temp:
if not temp.right and temp.left:
temp.right = TreeNode(None)
if temp.right and not temp.left:
temp.left = TreeNode(None)
stack.append(temp)
temp = temp.left
cur = stack.pop()
temp_list.append(cur.val)
temp = cur.right
return temp_list == temp_list[::-1]
def isSymmetric2(self, root):
"""这个方法适合满二叉树,获取是空节点补TreeNode()的树
:type root: TreeNode
:rtype: bool
"""
return self.preSearch(root) == self.postSearch(root)
def preSearch1(self, root):
temp = root
stack = list()
temp_list = [] # 几率元素个数
while temp or stack:
while temp:
temp_list.append(temp.val)
stack.append(temp)
temp = temp.left
cur = stack.pop()
temp = cur.right
return temp_list
def postSearch1(self, root):
temp = root
stack = list()
temp_list = [] # 几率元素个数
while temp or stack:
while temp:
temp_list.append(temp.val)
stack.append(temp)
temp = temp.right
cur = stack.pop()
temp = cur.left
return temp_list
def add_node(self, root):
if not root.right and root.left:
root.right = TreeNode()
if not root.left and root.right:
root.left = TreeNode()
if root.left:
self.add_node(root.left)
if root.right:
self.add_node(root.right)
def preSearch(self, root):
print(root.val)
if root.left:
self.preSearch(root.left)
if root.right:
self.preSearch(root.right)
def midSearch(self, root):
if root.left:
self.midSearch(root.left)
print(root.val)
if root.right:
self.midSearch(root.right)
def postSearch(self, root):
if root.left:
self.postSearch(root.left)
if root.right:
self.postSearch(root.right)
print(root.val)
def isSymmetric3(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
return self.check(root, root)
def check(self, left, right):
if not left and not right:
return True
if not left or not right:
return False
return left.val == left.val and self.check(left.left, right.right) and self.check(left.right, right.left)
def isSymmetric(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
queue = list()
queue.append(root)
queue.append(root)
while queue:
left = queue.pop()
right = queue.pop()
if not left and not right:
continue
if (not left or not right) or (left.val != right.val):
return False
queue.append(left.left)
queue.append(right.right)
queue.append(left.right)
queue.append(right.left)
return True
if __name__ == '__main__':
root = TreeNode(1)
n1 = TreeNode(2)
n2 = TreeNode(2)
n3 = TreeNode(3)
# n4 = TreeNode(4)
n5 = TreeNode(3)
# n6 = TreeNode(4)
n1.left = n3
# n1.right = n4
n2.left = n5
# n2.right = n6
root.left = n1
root.right = n2
s1 = Solution()
print(s1.isSymmetric(root))