112. Path Sum

1. Description

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.

2. Example

Example 1

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

Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There are 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.

3. Constraints

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

4. Solutions

Depth-first Search

n is the number of nodes in root
Time complexity: O(n)
Space complexity: O(logn->n)

class Solution {
public:
    bool hasPathSum(TreeNode *root, int targetSum) {
        if (root == nullptr) {
            return false;
        }

        if (root->left == nullptr && root->right == nullptr) {
            return root->val == targetSum;
        }

        int updated_target_sum = targetSum - root->val;
        return hasPathSum(root->left, updated_target_sum) ||
            hasPathSum(root->right, updated_target_sum);
    }
};