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113.path-sum-ii.py
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113.path-sum-ii.py
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#
# @lc app=leetcode id=113 lang=python3
#
# [113] Path Sum II
#
# https://leetcode.com/problems/path-sum-ii/description/
#
# algorithms
# Medium (56.90%)
# Likes: 6834
# Dislikes: 139
# Total Accepted: 730K
# Total Submissions: 1.3M
# Testcase Example: '[5,4,8,11,null,13,4,7,2,null,null,5,1]\n22'
#
# Given the root of a binary tree and an integer targetSum, return all
# root-to-leaf paths where the sum of the node values in the path equals
# targetSum. Each path should be returned as a list of the node values, not
# node references.
#
# A root-to-leaf path is a path starting from the root and ending at any leaf
# node. A leaf is a node with no children.
#
#
# Example 1:
#
#
# Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
# Output: [[5,4,11,2],[5,8,4,5]]
# Explanation: There are two paths whose sum equals targetSum:
# 5 + 4 + 11 + 2 = 22
# 5 + 8 + 4 + 5 = 22
#
#
# Example 2:
#
#
# Input: root = [1,2,3], targetSum = 5
# Output: []
#
#
# Example 3:
#
#
# Input: root = [1,2], targetSum = 0
# Output: []
#
#
#
# Constraints:
#
#
# The number of nodes in the tree is in the range [0, 5000].
# -1000 <= Node.val <= 1000
# -1000 <= targetSum <= 1000
#
#
#
# @lc code=start
# 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 pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
self.res = []
if not root:
return []
self.pathSumHelper(root, targetSum, [])
return self.res
def pathSumHelper(self, root, targetSum, path):
if not root:
return
if not root.left and not root.right:
if root.val == targetSum:
self.res.append(path + [root.val])
return
self.pathSumHelper(root.left, targetSum - root.val, path + [root.val])
self.pathSumHelper(root.right, targetSum - root.val, path + [root.val])
# @lc code=end