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,它的每个结点都存放着一个整数值。
找出路径和等于给定数值的路径总数。
路径不需要从根节点开始,也不需要在叶子节点结束,但是路径方向必须是向下的(只能从父节点到子节点)。
二叉树不超过1000个节点,且节点数值范围是 [-1000000,1000000] 的整数。
示例:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/
5 -3
/
3 2 11
/
3 -2 1
返回 3。和等于 8 的路径有:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
可使用以下main函数:
#include <iostream>
#include <queue>
#include <cstdlib>
#include <cstring>
#include<map>
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();
int sum;
cin>>sum;
int res=Solution().pathSum(root,sum);
cout<<res<<endl;
}
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。
最后输入一个整数sum。
输出说明 :
输出一个整数,表示结果。
输入范例 :
输出范例 :
#include <iostream> #include <queue> #include <cstdlib> #include <cstring> #include<map> 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: int res=0; void dfs(TreeNode* root, int sum) { if(root==NULL) return ; sum-=root->val; if(sum==0) res++; dfs(root->left,sum); dfs(root->right,sum); } int pathSum(TreeNode* root, int sum) { if(root==NULL) return res; dfs(root,sum); pathSum(root->left,sum); pathSum(root->right,sum); return res; } }; 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(); int sum; cin>>sum; int res=Solution().pathSum(root,sum); cout<<res<<endl; }