判断二叉树是否对称

判断一棵二叉树是否对称。二叉树节点定义如上次的结构相同：

typedef struct node {

int x;

struct node* left;

struct node* right;

} BN;

int isSymmetric(BN* root)；

来判断一棵二叉树是否对称，对称返回1，非对称返回0.

node结构要按照上面的代码在symmetric.h中进行定义。

注意被测试二叉树不一定是满二叉树。如果树不存在（根节点指针为空）返回0。如果除根节点外没有任何的其他节点返回1

#include<stdio.h>
#include<stdlib.h>
#define MAXD 50

typedef struct node {
int x;
struct node* left;
struct node* right;
} BN;

//建树
void buildTree(BN** root) {
int temp;
BN** que[MAXD];
int head = 0;
int tail = 1;
que[0] = root;
scanf("%d", &temp);
while (temp != -1) {
*que[head] = malloc(sizeof(struct node));
(*que[head])->x = temp;
que[tail] = &((*que[head])->left);
tail = (tail + 1) % MAXD;
que[tail] = &((*que[head])->right);
tail = (tail + 1) % MAXD;
head = (head + 1) % MAXD;
scanf("%d", &temp);
}
while (head != tail) {
*que[head] = NULL;
head = (head + 1) % MAXD;
}
}
void freeTree(BN* root) {
if (root != NULL) {
freeTree(root->left);
freeTree(root->right);
free(root);
}
}
//判断二叉树是否对称
int isSy2(BN *nodel, BN *noder) {
if (nodel == NULL && noder == NULL) {
return 1;
}
if (nodel != NULL && noder != NULL && noder->x == nodel->x) {
return isSy2(nodel->left, noder->right)&&isSy2(nodel->right, noder->left);
} else {
return 0;
}
}

int isSymmetric(BN* Prtroot) {
if (Prtroot == NULL) {
return 0;
}
return isSy2(Prtroot->left, Prtroot->right);
}
int main() {
BN* root = NULL;
buildTree(&root);
printf("%d\n", isSymmetric(root));
freeTree(root);
return 0;
}