symmetric-tree

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? > read more on how binary tree is serialized on OJ.

OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.
Here’s an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as”{1,2,3,#,#,4,#,#,5}”.

程序：

class Solution {public:    bool isSymmetric(TreeNode *root) {        if(root==NULL)            return true;        bool flag=true;        solve(flag,root->left,root->right);        return flag;    }    void solve(bool &flag,TreeNode *leftnode,TreeNode *rightnode)    {        if((leftnode&&rightnode==NULL)||(leftnode==NULL&&rightnode))        {            flag=false;            return;        }        if(leftnode==NULL&&rightnode==NULL)            return;        if(leftnode->val==rightnode->val)        {            solve(flag, leftnode->right,rightnode->left);            solve(flag, leftnode->left,rightnode->right);        }        else        {            flag=false;            return;        }    }};