LeetCode(Symmetric Tree)判断二叉树是否是对称的

题目要求：

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

可以用递归和非递归方式完成，非递归要比递归更容易，效率更高
代码：

class Solution {
public:

bool isSymmetric(TreeNode *root) {
if(root == NULL)
return true;

return DFS(root->left, root->right);
}

bool DFS(TreeNode* left, TreeNode* right)
{
if(left == NULL)
return (right == NULL);
if(right == NULL)
return (left == NULL);
if(left->val != right->val)
return false;
if(!DFS(left->left, right->right))
return false;
if(!DFS(left->right, right->left))
return false;
return true;
}

bool NonRecursive(TreeNode* root)
{
if(root == NULL)
return true;
queue<TreeNode*> q1;
queue<TreeNode*> q2;
q1.push(root->left);
q2.push(root->right);
while(!q1.empty() && !q2.empty())
{
TreeNode* node1 = q1.front();
TreeNode* node2 = q2.front();
q1.pop();
q2.pop();
if((node1 == NULL && node2 != NULL) || (node1 != NULL && node2 == NULL))
return false;
if(node1 != NULL && node2 != NULL)
{
if(node1->val != node2->val)
return false;
q1.push(node1->left);
q1.push(node1->right);
q2.push(node2->right);
q2.push(node2->left);
}
}
return q1.empty() && q2.empty();
}

};