/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> res;
vector<int> tmp;
if (root == NULL)
return res;
hasPathSum(root, 0, sum, res, tmp);
return res;
}
void hasPathSum(TreeNode* root, int sum, int target, vector<vector<int>>& res, vector<int>& tmp){
if (root == NULL)
return;
if (root->left == NULL && root->right == NULL && sum + root->val == target){
tmp.push_back(root->val);
res.push_back(tmp);
tmp.pop_back();
return;
}
tmp.push_back(root->val);
hasPathSum(root->left, root->val + sum, target, res, tmp);
hasPathSum(root->right, root->val + sum, target, res, tmp);
tmp.pop_back();
}
};
static int x=[](){
std::ios::sync_with_stdio(false);
cin.tie(NULL);
return 0;
}();
LetCode 113. 路径总和 II
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转载自blog.csdn.net/wbb1997/article/details/81089406
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