Date and Time: Jul 16, 2024, 21:17 (EST)
Link: https://leetcode.com/problems/path-sum/
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 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.
Edge case:
Input: root = [1, 2], targetSum = 1
Output: false
Explanation: We want the tree has a root-to-leaf path.
-
The number of nodes in the tree is in the range
[0, 5000]. -
-1000 <= Node.val <= 1000 -
-1000 <= targetSum <= 1000
We recurse dfs on root.left and root.right, until we reach the leaf (if not root.left and not root.right), we start comparing if curSum == targetSum.
# 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:
# Running DFS from root to leaf, return True if curSum == targetSum
# TC: O(n), n is # nodes in the tree, SC: O(1)
def dfs(node, curSum):
if not node:
return False
curSum += node.val
# When we reach the leaf node
if not node.left and not node.right:
return curSum == targetSum
return dfs(node.left, curSum) or dfs(node.right, curSum)
return dfs(root, 0)Time Complexity:
Space Complexity:
