LeetCode-101. Symmetric Tree-using C

【题目描述】
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即判断一棵二叉树是否是对称的
【函数形式】/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool isSymmetric(struct TreeNode* root) {

}
【解题思路】
左子树为空，右子树也为空，或者，左子树与右子树的value相等时，以根节点为根的二叉树是对称的。那么我们可以设计这样一个函数，传入两个指向树节点的指针，判断两个指针所指向的结点是否对称，从根节点开始递归调用该函数。
【代码如下】bool function(struct TreeNode* left, struct TreeNode* right);
bool isSymmetric(struct TreeNode* root) {
return function(root,root);
}
bool function(struct TreeNode* left, struct TreeNode* right)
{
if(left == NULL && right == NULL)
return 1;
if (left->val != right->val)
return 0;
if(left == NULL && right != NULL)
return 0;
if(left != NULL && right == NULL)
return 0;

int r1,r2;

if(left->left != NULL && right->right != NULL)
{
r1 = function( (left->left), (right->right));
}
else if (left->left == NULL && right->right == NULL)
{
r1 = 1;
}
else
{
return 0;
}

if (left == right)
{
//root
return r1;
}

if(right->left != NULL && left->right != NULL)
{
r2 = function( (left->right), (right->left));
}
else if (right->left == NULL && left->right == NULL)
{
r2 = 1;
}
else
{
return 0;
}

return r1&&r2;
}