问题描述:
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3]
is symmetric:
1 / 2 2 / / 3 4 4 3
But the following [1,2,2,null,3,null,3]
is not:
1 / 2 2 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
思路:二分类树的问题,可以考虑递归解法
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def isSymmetric(self, root: TreeNode) -> bool: if root == None: return True def ismirror(root1,root2): if root1 == None and root2 ==None: return True elif root1 == None or root2 ==None or root1.val != root2.val: return False return ismirror(root1.left,root2.right) and ismirror(root1.right,root2.left) return ismirror(root.left,root.right)