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题目：
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following is not:
1 / 2 2 \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}" means?
解题思路：
将根节点的左右孩子分开当做两个树来处理，注意它是要几何对称，所以左边的树的左孩子和右边树的右孩子相互比较。
1 class Solution(object): 2 def isSymmetric(self, root): 3 """ 4 :type root: TreeNode 5 :rtype: bool 6 """ 7 from collections import deque 8 9 if not root: 10 return True 11 12 p = root.left 13 q = root.right 14 15 if p == None and q != None: 16 return False 17 elif p != None and q == None: 18 return False 19 elif p == None and q == None: 20 return True 21 else: 22 pqueue = deque([p]) 23 qqueue = deque([q]) 24 25 while pqueue and qqueue: 26 nodep = pqueue.popleft() 27 nodeq = qqueue.popleft() 28 29 if nodep.val != nodeq.val: 30 return False 31 else: 32 if nodep.left and nodeq.right: 33 pqueue.append(nodep.left) 34 qqueue.append(nodeq.right) 35 elif nodep.left == None and nodeq.right != None: 36 return False 37 elif nodep.left != None and nodeq.right == None: 38 return False 39 else: 40 pass 41 42 if nodep.right and nodeq.left: 43 pqueue.append(nodep.right) 44 qqueue.append(nodeq.left) 45 elif nodep.right == None and nodeq.left != None: 46 return False 47 elif nodep.right != None and nodeq.left == None: 48 return False 49 else: 50 pass 51 52 return True 53
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原文地址：http://www.cnblogs.com/tingyu-yudeyinji/p/4771023.html
