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 and
sum = 22
,
5
/
4 8
/ /
11 13 4
/ /
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]
解题思路:
通过遍历树保存从根到叶子节点的路径和路径和。然后推断其和是否等于sum就可以.
解题代码:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void dfs(TreeNode *rt,long long sum,vector<vector<int> > &ans,vector<int> &path)
{
path.push_back(rt->val);
if (rt->left == rt->right && rt->left == NULL)
{
if (sum == rt->val)
ans.push_back(vector<int>(path.begin(),path.end()));
path.erase(path.end()-1);
return ;
}
if (rt->left)
dfs(rt->left,sum - rt->val,ans,path);
if (rt->right)
dfs(rt->right,sum - rt->val,ans,path);
path.erase(path.end()-1);
}
vector<vector<int> > pathSum(TreeNode *root, int sum)
{
vector<vector<int> > ans ;
if (!root)
return ans ;
vector<int> path;
dfs(root,sum,ans,path);
return ans ;
}
};