Path Sum II Leetcode Solution

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.

Path Sum II Leetcode Solution
Path Sum II Leetcode Solution

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:

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, lets 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.

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