劍指offer之中判斷二叉樹是不是對稱二叉樹（遞歸和非遞歸實現）

1 問題

判斷二叉樹是不是對稱（遞歸和非遞歸實現）

如下二叉樹，就是對稱的二叉樹

	          2
               3     3           
             1  4   4  1 

如下二叉樹，就是非對稱的二叉樹 

	          2
               3     3           
             1  4   4  2

 

 

 

 

 

2 代碼實現

#include <iostream>
#include <queue>

using namespace std;

#define true 1
#define false 0

typedef 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);

/*
 *判斷是否是對稱二叉樹（遞歸實現）
 */
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);
} 

/*
 *判斷是否是對稱二叉樹（非遞歸實現）
 */
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;
		}	
		//因爲上面情況只包含left爲NULL和right不爲NULL以及left不爲NULL和right爲NULL，
		//還包含2種情況，left和right都爲NULL,以及left和right都不爲NULL,所以我們left->value和right->判斷相等的時候
		//一定要記得對left和right都不是NULL的前提下才能調用->,下次切記，看到指針->的時候需要判斷指針是否爲NULL
		//left和right都爲NULL的時候，我們直接對queue1和queue2進行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;
}

 

 

 

3 運行結果

tree is symmertric
tree is symmertric