112. Path Sum
Easy LeetCodeGiven 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)
Python
# 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)
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