Path Sum II LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
Share:

Path Sum II 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

Path Sum II Solutions

Time: O(n log n )
Space: O(n)

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();
  }
};

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);
  }
}

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

Watch Tutorial

Checkout more Solutions here

Leave a Comment

Your email address will not be published. Required fields are marked *

x