力扣练习——41 对称二叉树

41 对称二叉树

1.问题描述
给定一个二叉树，检查它是否是镜像对称的。

例如，二叉树 [1,2,2,3,4,4,3] 是对称的。

1

/ \

2 2

/ \ / \

3 4 4 3

二叉树[1,2,2,null,3,3,null,5,-2,-2,5]是对称的。

                    1

                /        \

              2           2

               \         /

                3      3

              /  \    /  \

            5  -2  -2  5

但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:

1

/ \

2 2

\ \

3 3

可使用以下main函数：

#include

#include

#include

#include

using namespace std;

struct TreeNode

{

int val;

TreeNode *left;

TreeNode *right;

TreeNode() : val(0), left(NULL), right(NULL) {}

TreeNode(int x) : val(x), left(NULL), right(NULL) {}

TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}

};

TreeNode* inputTree()

{

int n,count=0;

char item[100];

cin>>n;

if (n==0)

    return NULL;

cin>>item;

TreeNode* root = new TreeNode(atoi(item));

count++;

queue<TreeNode*> nodeQueue;

nodeQueue.push(root);

while (count<n)

{

    TreeNode* node = nodeQueue.front();

    nodeQueue.pop();

    cin>>item;

    count++;

    if (strcmp(item,"null")!=0)

    {

        int leftNumber = atoi(item);

        node->left = new TreeNode(leftNumber);

        nodeQueue.push(node->left);

    }

    if (count==n)

        break;

    cin>>item;

    count++;

    if (strcmp(item,"null")!=0)

    {

        int rightNumber = atoi(item);

        node->right = new TreeNode(rightNumber);

        nodeQueue.push(node->right);

    }

}

return root;

}

int main()

{

TreeNode* root;

root=inputTree();

bool res=Solution().isSymmetric(root);

cout<<(res?“true”:“false”);

}

2.输入说明
首先输入结点的数目n（注意，这里的结点包括题中的null空结点）

然后输入n个结点的数据，需要填充为空的结点，输入null。
3.输出说明
输出true或false
4.范例
输入
11
1 2 2 null 3 3 null 5 -2 -2 5
输出
true
5.代码

#include <iostream>
#include <queue>
#include <cstdlib>
#include <cstring>
using namespace std;

struct TreeNode

{
    

	int val;

	TreeNode *left;

	TreeNode *right;

	TreeNode() : val(0), left(NULL), right(NULL) {
    }

	TreeNode(int x) : val(x), left(NULL), right(NULL) {
    }

	TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {
    }

};

TreeNode* inputTree()

{
    
	int n, count = 0;
	char item[100];
	cin >> n;
	if (n == 0)
		return NULL;

	cin >> item;

	TreeNode* root = new TreeNode(atoi(item));

	count++;

	queue<TreeNode*> nodeQueue;

	nodeQueue.push(root);

	while (count < n)

	{
    

		TreeNode* node = nodeQueue.front();

		nodeQueue.pop();

		cin >> item;

		count++;

		if (strcmp(item, "null") != 0)

		{
    

			int leftNumber = atoi(item);

			node->left = new TreeNode(leftNumber);

			nodeQueue.push(node->left);

		}

		if (count == n)

			break;

		cin >> item;

		count++;

		if (strcmp(item, "null") != 0)

		{
    

			int rightNumber = atoi(item);

			node->right = new TreeNode(rightNumber);

			nodeQueue.push(node->right);

		}

	}

	return root;

}


bool cmp(TreeNode *tree1, TreeNode *tree2)
{
    
	//1.两棵树均为空
	if (tree1 == NULL && tree2 == NULL)
		return true;
	//2.有一棵树非空或者两个根节点值不相等
	if (tree1 == NULL || tree2 == NULL || tree1->val != tree2->val)
		return false;
	//3.比较左子树的右节点和右子树的左节点，左子树的左节点和右子树的右节点
	return cmp(tree1->left, tree2->right) && cmp(tree1->right, tree2->left);
}

bool isSymmetric(TreeNode *root)
{
    
	//1.树为空
	if (root == NULL)
		return true;
	//2.判断左右子树
	return cmp(root->left, root->right);
}


int main()

{
    

	TreeNode* root;

	root = inputTree();

	bool res = isSymmetric(root);

	cout << (res ? "true" : "false")<<endl;

}