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113. Path Sum II

Difficulty: Medium

Related Topics: Backtracking, Tree, DFS, Binary Tree

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: []

First Attempt:

Code:

# 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]]:
        temp = []
        ans = []
        sam = 0
        dup = root
        def dfs(root, targetSum, sam, temp):
            if root:
                if root.left == None and root.right == None:
                    if sam != targetSum:
                        return False
                    else:
                        return True
                if root.left:
                    temp.append(root.val)
                    if dfs(root.left, targetSum, sam+root.left.val, temp):     
                        temp.append(root.left.val)
                        ans.append(temp.copy())
                    else:
                        temp.pop()
                if root.right:
                    temp.append(root.val)
                    if dfs(root.right, targetSum, sam+root.right.val, temp):
                        temp.append(root.right.val)
                        ans.append(temp.copy())
                        temp=[]
                    else:
                        temp.pop()
                        
        dfs(dup, targetSum, root.val, temp)
        return ans

Second Attempt:

Code:

class Solution:
    def pathSum(self, root, targetSum):
        result = []
        def dfs(root, path):
            nonlocal result
            if not root:
                return []
            path += [root.val]           
            if not root.left and not root.right:
			    # Adding the condition needed in the question
                if sum(path)==targetSum:
                    result.append(path.copy())
            dfs(root.left, path)
            dfs(root.right, path)
            path.pop()
            return result    
        return dfs(root, [])

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