## Path Sum II

• Time:O(n\log n)
• Space:O(n)

## C++

``````class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> ans;
dfs(root, sum, {}, ans);
return ans;
}

private:
void dfs(TreeNode* root, int sum, vector<int>&& path,
vector<vector<int>>& ans) {
if (!root)
return;
if (root->val == sum && !root->left && !root->right) {
path.push_back(root->val);
ans.push_back(path);
path.pop_back();
return;
}

path.push_back(root->val);
dfs(root->left, sum - root->val, move(path), ans);
dfs(root->right, sum - root->val, move(path), ans);
path.pop_back();
}
};
``````

## JAVA

``````class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> ans = new ArrayList<>();
dfs(root, sum, new ArrayList<>(), ans);
return ans;
}

private void dfs(TreeNode root, int sum, List<Integer> path, List<List<Integer>> ans) {
if (root == null)
return;
if (root.val == sum && root.left == null && root.right == null) {
path.remove(path.size() - 1);
return;
}

dfs(root.left, sum - root.val, path, ans);
dfs(root.right, sum - root.val, path, ans);
path.remove(path.size() - 1);
}
}
``````

## Python

``````class Solution:
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
ans = []

def dfs(root: TreeNode, sum: int, path: List[int]) -> None:
if root is None:
return
if root.val == sum and root.left is None and root.right is None:
ans.append(path + [root.val])
return

dfs(root.left, sum - root.val, path + [root.val])
dfs(root.right, sum - root.val, path + [root.val])

dfs(root, sum, [])
return ans
``````