作者：我从不在乎O心痛 | 来源：互联网 | 2023-10-10 22:03
Givenabinarytreeandasum,findallroot-to-leafpathswhereeachpathssumequalsthegivensu
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree andsum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]
题意：给定一数，在树中找出所有路径和等于该数的情况。
方法一：
使用vector向量实现stack的功能，以方便输出指定路径。思想和代码和Path sum大致相同。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vectorint> > pathSum(TreeNode *root, int sum)
{
vectorint>> res;
vector vec;
TreeNode *pre=NULL;
TreeNode *cur=root;
int temVal=0;
while(cur|| !vec.empty())
{
while(cur)
{
vec.push_back(cur);
temVal+=cur->val;
cur=cur->left;
}
cur=vec.back();
if(cur->left==NULL&&cur->right==NULL&&temVal==sum)
{ //和Path sum最大的区别
vector<int> temp;
for(int i=0;ii)
temp.push_back(vec[i]->val);
res.push_back(temp);
}
if(cur->right&&cur->right !=pre)
cur=cur->right;
else
{
vec.pop_back();
temVal-=cur->val;
pre=cur;
cur=NULL;
}
}
return res;
}
};
方法二：递归法
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vectorint> > pathSum(TreeNode *root, int sum)
{
vectorint>> res;
vector<int> path;
findPaths(root,sum,res,path);
return res;
}
void findPaths(TreeNode *root,int sum,vectorint>> &res,vector<int> &path)
{
if(root==NULL) return;
path.push_back(root->val);
if(root->left==NULL&&root->right==NULL&&sum==root->val)
res.push_back(path);
findPaths(root->left,sum-root->val,res,path);
findPaths(root->right,sum-root->val,res,path);
path.pop_back();
}
};
京公网安备 11010802041100号