作者:
晚于: 2020-08-05 12:00:00后提交分数乘系数50%
截止日期: 2020-08-12 12:00:00
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 <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;
}
int main()
{
TreeNode* root;
root=inputTree();
bool res=Solution().isSymmetric(root);
cout<<(res?"true":"false");
}
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。
输出说明 :
输出true或false
输入范例 :
输出范例 :
true
#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) {} }; class Solution { public: bool judge(TreeNode* l,TreeNode* r) { if(!l&&!r) return true; if(!l||!r) return false; return l->val==r->val&&judge(l->right,r->left) &&judge(l->left,r->right); } bool isSymmetric(TreeNode* root) { return judge(root,root) ; } }; 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"); }