对称二叉树

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• 递归解法：

# Definition for a binary tree node.
# 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
        def dfs(left,right):
            if left is None and right is None:
                return True
            elif not (left and right):
                return False
            elif left.val == right.val:
                # 是关键，对称的条件：左子树的左节点和右子树的右节点、左子树右节点与右子树左节点要相同
                return dfs(left.left,right.right) and dfs(left.right,right.left)
            else:
                return False
        return dfs(root.left,root.right)