/**
* 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;
if(root == NULL) return res;
vector<int> add;
add.push_back(root->val);
DFS(res,add,root,sum,root->val);
return res; }
void DFS(vector<vector<int>>& res,vector<int>& add,TreeNode*root,int sum,int& he){
if((root->left == NULL)&&(root->right == NULL)){
if(he == sum) res.push_back(add);
}
else if((root->left != NULL)&&(root->right == NULL)){
add.push_back(root->left->val);
he += root->left->val;
DFS(res,add,root->left,sum,he);
add.pop_back();
he -= root->left->val;
}
else if((root->left == NULL)&&(root->right != NULL)){
add.push_back(root->right->val);
he += root->right->val;
DFS(res,add,root->right,sum,he);
add.pop_back();
he -= root->right->val;
}
else if((root->left != NULL)&&(root->right != NULL)){
add.push_back(root->left->val);
he += root->left->val;
DFS(res,add,root->left,sum,he);
add.pop_back();
he -= root->left->val;
add.push_back(root->right->val);
he += root->right->val;
DFS(res,add,root->right,sum,he);
add.pop_back();
he -= root->right->val;
}
return;
}
};

_虽然代码丑,但比较好理解

class Solution {
public:
vector<vector<int> > pathSum(TreeNode *root, int sum) {
vector<vector<int>> res;
vector<int> out;
helper(root, sum, out, res);
return res;
}
void helper(TreeNode* node, int sum, vector<int>& out, vector<vector<int>>& res) {
if (!node) return;
out.push_back(node->val);
if (sum == node->val && !node->left && !node->right) {
res.push_back(out);
}
helper(node->left, sum - node->val, out, res);
helper(node->right, sum - node->val, out, res);
out.pop_back();
}
};

——这个和上一题对应

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