Question：

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

判断二叉树是否左右对称

Algorithm：

对左右子树BFS（宽度优先搜索），要注意判断左右子树一个非空一个不为空的情况

Accepted Code：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root==NULL)return true;
        queue<TreeNode*> l;   //左子树
        queue<TreeNode*> r;   //右子树
        if(root->left==NULL&&root->right==NULL)
            return true;
        else if(root->left!=NULL&&root->right!=NULL)
        {
            l.push(root->left);
            r.push(root->right);
        }
        else
            return false;
        while(!l.empty()&&!r.empty())
        {
            TreeNode* temp1=l.front();
            TreeNode* temp2=r.front();
            if(temp1->val!=temp2->val)
                return false;
            l.pop();
            r.pop();
            if(temp1->left&&temp2->right)
            {
                l.push(temp1->left);
                r.push(temp2->right);
            }
            else if((temp1->left==NULL^temp2->right==NULL)==1)
                return false;
            if(temp1->right&&temp2->left)
            {
                l.push(temp1->right);
                r.push(temp2->left);
            }
            else if((temp1->right==NULL^temp2->left==NULL)==1)
                return false;
        }
        if(l.empty()&&r.empty())
            return true;
        else
            return false;
    }
};