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本文为senlie原创，转载请保留此地址：http://blog.csdn.net/zhengsenlie
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? >
read more on how binary tree is serialized on OJ.
题意：给定一棵二叉树，判断它是否对称
思路：dfs递归
一棵二叉树对称的条件：
1.左右子树的根节点一样
2.左子树的左子树和右子树的右子树对称
3.左子树的右子树和右子树的左子树对称
复杂度：时间O(n) ，空间O(log n)
相关题目：Same Tree
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode *left, TreeNode *right)
{
if(!left && !right ) return true;
if(!left || !right) return false;
if(left->val != right->val) return false;
return (isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left));
}
bool isSymmetric(TreeNode *root){
if(!root) return true;
return isSymmetric(root->left, root->right);
}
};Leetcode 树 Symmetric Tree,布布扣,bubuko.com
标签：style blog class code c tar
原文地址：http://blog.csdn.net/zhengsenlie/article/details/25630467
