Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where each path's sum equals targetSum.
A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 Output: [[5,4,11,2],[5,8,4,5]]
Example 2:
Input: root = [1,2,3], targetSum = 5 Output: []
Example 3:
Input: root = [1,2], targetSum = 0 Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 5000]. -1000 <= Node.val <= 1000-1000 <= targetSum <= 1000
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
def dfs(root, sum):
if root is None:
return
path.append(root.val)
if root.val == sum and root.left is None and root.right is None:
res.append(path.copy())
dfs(root.left, sum - root.val)
dfs(root.right, sum - root.val)
path.pop()
if not root:
return []
res = []
path = []
dfs(root, sum)
return res/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private List<List<Integer>> res;
private List<Integer> path;
public List<List<Integer>> pathSum(TreeNode root, int sum) {
if (root == null) return Collections.emptyList();
res = new ArrayList<>();
path = new ArrayList<>();
dfs(root, sum);
return res;
}
private void dfs(TreeNode root, int sum) {
if (root == null) return;
path.add(root.val);
if (root.val == sum && root.left == null && root.right == null) {
res.add(new ArrayList<>(path));
}
dfs(root.left, sum - root.val);
dfs(root.right, sum - root.val);
path.remove(path.size() - 1);
}
}