# 113. Path Sum II

## Description

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.

A root-to-leaf path is a path starting from the root and ending at any leaf node. 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]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22


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

## Solutions

Solution 1: DFS

We start from the root node, recursively traverse all paths from the root node to the leaf nodes, and record the path sum. When we traverse to a leaf node, if the current path sum equals targetSum, then we add this path to the answer.

The time complexity is $O(n^2)$, where $n$ is the number of nodes in the binary tree. The space complexity is $O(n)$.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private List<List<Integer>> ans = new ArrayList<>();
private List<Integer> t = new ArrayList<>();

public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
dfs(root, targetSum);
return ans;
}

private void dfs(TreeNode root, int s) {
if (root == null) {
return;
}
s -= root.val;
if (root.left == null && root.right == null && s == 0) {
}
dfs(root.left, s);
dfs(root.right, s);
t.remove(t.size() - 1);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
vector<vector<int>> ans;
vector<int> t;
function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int s) {
if (!root) return;
s -= root->val;
t.emplace_back(root->val);
if (!root->left && !root->right && s == 0) ans.emplace_back(t);
dfs(root->left, s);
dfs(root->right, s);
t.pop_back();
};
dfs(root, targetSum);
return ans;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
def dfs(root, s):
if root is None:
return
s += root.val
t.append(root.val)
if root.left is None and root.right is None and s == targetSum:
ans.append(t[:])
dfs(root.left, s)
dfs(root.right, s)
t.pop()

ans = []
t = []
dfs(root, 0)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func pathSum(root *TreeNode, targetSum int) (ans [][]int) {
t := []int{}
var dfs func(*TreeNode, int)
dfs = func(root *TreeNode, s int) {
if root == nil {
return
}
s -= root.Val
t = append(t, root.Val)
if root.Left == nil && root.Right == nil && s == 0 {
ans = append(ans, slices.Clone(t))
}
dfs(root.Left, s)
dfs(root.Right, s)
t = t[:len(t)-1]
}
dfs(root, targetSum)
return
}

• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} targetSum
* @return {number[][]}
*/
var pathSum = function (root, targetSum) {
const ans = [];
const t = [];
function dfs(root, s) {
if (!root) return;
s -= root.val;
t.push(root.val);
if (!root.left && !root.right && s == 0) ans.push([...t]);
dfs(root.left, s);
dfs(root.right, s);
t.pop();
}
dfs(root, targetSum);
return ans;
};


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(
root: Option<Rc<RefCell<TreeNode>>>,
paths: &mut Vec<i32>,
mut target_sum: i32,
res: &mut Vec<Vec<i32>>
) {
if let Some(node) = root {
let mut node = node.borrow_mut();
target_sum -= node.val;
paths.push(node.val);
if node.left.is_none() && node.right.is_none() {
if target_sum == 0 {
res.push(paths.clone());
}
} else {
if node.left.is_some() {
Self::dfs(node.left.take(), paths, target_sum, res);
}
if node.right.is_some() {
Self::dfs(node.right.take(), paths, target_sum, res);
}
}
paths.pop();
}
}

pub fn path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> Vec<Vec<i32>> {
let mut res = vec![];
let mut paths = vec![];
Self::dfs(root, &mut paths, target_sum, &mut res);
res
}
}