小知识，大挑战！本文正在参与“程序员必备小知识”创作活动。
描述
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
Input: root = [1,2,2,3,4,4,3]
Output: true
Example 2:
Input: root = [1,2,2,null,3,null,3]
Output: false
Note:
The number of nodes in the tree is in the range [1, 1000].
-100 <= Node.val <= 100
解析
根据题意，给出了一个二叉树，判断这个树是不是根据中线对称的。一开始思路比较狭窄，光想着对树本身是用递归，发现很难，最后看到了大神的解法，清晰且容易。
思路比较简单，就是定义一个递归函数 symmetric ，有两个参数 left 和 right ，分别代表了两个树的根节点 在 symmetric 函数中传入两个相同的 root ，对两个相同的树进行操作，判断 left 的值是否和 right 的值是否相等，如果相等继续递归 symmetric(left.left, right.right) and symmetric(left.right, right.left) ，如果不相等直接返回 False 即可。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def isSymmetric(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
def symmetric(left, right):
if not left and not right: return True
if left and right and left.val == right.val:
return symmetric(left.left, right.right) and symmetric(left.right, right.left)
return False
return symmetric(root, root)
运行结果
Runtime: 20 ms, faster than 84.75% of Python online submissions for Symmetric Tree.
Memory Usage: 13.6 MB, less than 43.91% of Python online submissions for Symmetric Tree.
解析
当然同样可以使用栈来迭代树中的节点来进行比较。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def isSymmetric(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if not root:
return True
stack = [(root.left, root.right)]
while stack:
l, r = stack.pop()
if not l and not r:
continue
if not l or not r or (l.val != r.val):
return False
stack.append((l.left, r.right))
stack.append((l.right, r.left))
return True
运行结果
Runtime: 40 ms, faster than 12.21% of Python online submissions for Symmetric Tree.
Memory Usage: 13.6 MB, less than 70.59% of Python online submissions for Symmetric Tree.
原题链接：leetcode.com/problems/sy…
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