剑指offer之中判断二叉树是不是对称二叉树（递归和非递归实现）

1 问题

判断二叉树是不是对称（递归和非递归实现）

如下二叉树，就是对称的二叉树

	          23     3           1  4   4  1

如下二叉树，就是非对称的二叉树

	          23     3           1  4   4  2

2 代码实现

#include <iostream>
#include <queue>using namespace std;#define true 1
#define false 0typedef struct Node
{int value;struct Node* left;struct Node* right;
} Node;int isSymmetricTree(Node *head);
int isSymmetric(Node *left, Node *right);
int isSymmetricTree1(Node *head);/**判断是否是对称二叉树（递归实现）*/
int isSymmetricTree(Node *head)
{if (head == NULL){return true;}   return isSymmetric(head, head);
}int isSymmetric(Node *left, Node *right)
{if (left == NULL && right == NULL){return true;}if (left == NULL || right == NULL){return false;}if (left->value != right->value){return false;}return isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left);
} /**判断是否是对称二叉树（非递归实现）*/
int isSymmetricTree1(Node *head)
{if (head == NULL){return true;}   std::queue<Node *> queue1;std::queue<Node *> queue2;queue1.push(head->left);queue2.push(head->right);while(!queue1.empty() || !queue2.empty()){Node *left = queue1.front();Node *right = queue2.front();if ((left != NULL) && (right == NULL)){return false;}if ((left == NULL) && (right != NULL)){return false;}	//因为上面情况只包含left为NULL和right不为NULL以及left不为NULL和right为NULL，//还包含2种情况，left和right都为NULL,以及left和right都不为NULL,所以我们left->value和right->判断相等的时候//一定要记得对left和right都不是NULL的前提下才能调用->,下次切记，看到指针->的时候需要判断指针是否为NULL//left和right都为NULL的时候，我们直接对queue1和queue2进行pop()操作if (left && right && (left->value != right->value)){return false;}queue1.pop();queue2.pop();if (left != NULL){queue1.push(left->left);queue1.push(left->right);}if (right != NULL){queue2.push(right->right);queue2.push(right->left);}}return true;
}int main()
{/*              2*           3     3           *         1  4   4  1        *       */Node head1, node1, node2, node3, node4, node5, node6;Node head2, node7, node8;head1.value = 2;node1.value = 3;node2.value = 3;node3.value = 1;node4.value = 4;node5.value = 4;node6.value = 1;head1.left = &node1;head1.right = &node2;node1.left = &node3;node1.right = &node4;node2.left = &node5;node2.right = &node6;node3.left = NULL;node3.right = NULL;node4.left = NULL;node4.right = NULL;node5.left = NULL;node5.right = NULL;node6.left = NULL;node6.right = NULL;if (isSymmetricTree(&head1)){std::cout << "tree is symmertric" << std::endl;}else{std::cout << "tree is not symmertric" << std::endl;}if (isSymmetricTree1(&head1)){std::cout << "tree is symmertric" << std::endl;}else{std::cout << "tree is not symmertric" << std::endl;}return 0;
}