Path Sum II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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 ProblemPath 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 references.

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:

pathsumii1
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:

pathsum2
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
Path Sum II– LeetCode Solutions
Path Sum II Solution in 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();
  }
};
Path Sum II 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);
  }
}
Path Sum II Solution 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

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