LC 112. Path Sum

Alen Alex · February 7, 2024

LeetCode Link: 112. Path Sum
Language: C#

Problem Statement

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Examples

Example 1:

img1
Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.

Example 2:

img1
Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 –> 2): The sum is 3.
(1 –> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.

Example 3:

Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.

Constraints

  • The number of nodes in the tree is in the range [0, 5000].
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution 
{
    public bool HasPathSum(TreeNode root, int targetSum) 
    {
        return HasSum(root, targetSum, 0);
    }

    private bool HasSum(TreeNode root, int target, int sum)
    {
        if (root is null)
        {
            return false;
        }

        sum += root.val;
        if (root.left is null && root.right is null)
        {
            return sum == target;
        }

        return HasSum(root.left, target, sum) || HasSum(root.right, target, sum);
    }
}

Complexity

Time Complexity: O(N)
Space Complexity: O(1)

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