112. Path Sum

Easy LeetCode

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.

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.

Constraints

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

How to solve the problem

Code

• Recursion (DFS)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        def dfs(node, current_sum):
            if not node:
                return False
            current_sum += node.val
            if not node.left and not node.right:
                return current_sum == targetSum
            return dfs(node.left, current_sum) or dfs(node.right, current_sum)

        return dfs(root, 0)

Complexity

• Time complexity: O(n), n == number of nodes
• Space complexity: O(n)