In this post, we are going to solve the Path Sum II Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Problem
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
Now, let’s see the leetcode solution of Path Sum II Leetcode Solution.
Path Sum II Leetcode Solution in Python
class Solution: def pathSum(self, root: TreeNode, summ: int) -> List[List[int]]: ans = [] def dfs(root: TreeNode, summ: int, path: List[int]) -> None: if not root: return if root.val == summ and not root.left and not root.right: ans.append(path + [root.val]) return dfs(root.left, summ - root.val, path + [root.val]) dfs(root.right, summ - root.val, path + [root.val]) dfs(root, summ, []) return ans
Path Sum II Leetcode Solution in CPP
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 == nullptr) return; if (root->val == sum && root->left == nullptr && root->right == nullptr) { 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(); } };
Path Sum II Leetcode Solution in 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.add(root.val); ans.add(new ArrayList<>(path)); path.remove(path.size() - 1); return; } path.add(root.val); dfs(root.left, sum - root.val, path, ans); dfs(root.right, sum - root.val, path, ans); path.remove(path.size() - 1); } }
Note: This problem Path Sum II is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purpose.