Path Sum

Published: Oct 4, 2022

Introduction

The pre-order traversal works to find the answer. When the depth-first search (recursion) approach is taken, we should be careful about the value consistency when the process comes back from deeper stack. Other than that, the problem is just a tree traversal.

Problem 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.

Constraints:

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

https://leetcode.com/problems/path-sum/

Examples

Example 1
Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation:
          5
        /   \
      4      8
    /      /   \
  11      13    4
 /  \            \
7    2            1
Path: 5 -> 4 -> 11 -> 2

Example 2
Input: root = [1,2,3], targetSum = 5
Output: false
Explanation:
   1
 /  \
2    3

Example 3
Input: root = [], targetSum = 0
Output: false
Explanation: The tree is empty.

Analysis

The solution here takes the pre-order traversal approach (depth-first search). The first step is to subtract the root value from the given targetSum. If the current node is a leaf, check the targetSum becomes zero. The left or right subtree returns true if the targetSum exists.

Solution

# 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 PathSum:
    def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        if not root:
            return False
        targetSum -= root.val
        if not root.left and not root.right:
            return targetSum == 0
        left = self.hasPathSum(root.left, targetSum)
        right = self.hasPathSum(root.right, targetSum)
        return left or right

Complexities

• Time: O(n)
• Space: O(1)