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In this post, you will find the solution for the **Path Sum II** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Path Sum II– LeetCode Problem

Path Sum II– LeetCode Problem

**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 referencesA **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 = 22Output:[[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 = 5Output:[]

**Example 3:**

Input:root = [1,2], targetSum = 0Output:[]

**Constraints:**

- The number of nodes in the tree is in the range
`[0, 5000]`

. `-1000 <= Node.val <= 1000`

`-1000 <= targetSum <= 1000`

Path Sum II– LeetCode Solutions

Path Sum II Solution in C++:

class Solution { public: vector> pathSum(TreeNode* root, int sum) { vector > ans; dfs(root, sum, {}, ans); return ans; } private: void dfs(TreeNode* root, int sum, vector && path, vector >& 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(); } };

Path Sum II Solution in Java:

class Solution { public List> pathSum(TreeNode root, int sum) { List

> ans = new ArrayList<>(); dfs(root, sum, new ArrayList<>(), ans); return ans; } private void dfs(TreeNode root, int sum, List

path, List > 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); } }

Path Sum IISolution in 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