113.Path_Sum_II(Medium).md

113. Path Sum II (Medium)

Date and Time: Nov 11, 2024, 16:12 (EST)

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

Question:

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

Walk-through:

1. Use curSum to keep track of current path's sum, stack[] to keep track of current path values.

2. We then run DFS from the root, when we reach the leaf node, we check if curSum == targetSum, if so, we save the copy of current stack[] into res[]. Then, we pop() the current node and decrement curSum -= node.val.

Python 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 Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        # Run DFS from root to leaf
        # Use curSum and stack[] to keep track of current path and its sum
        # When curSum == targetSum, copy current stack[] into res[]

        # TC: O(n), n is # of nodes in tree, SC: O(n)
        curSum, stack = 0, []
        res = []
        def dfs(node):
            nonlocal curSum, stack, res
            if not node:
                return
            curSum += node.val
            stack.append(node.val)
            # When we reach the leaf and curSum == targetSum
            if not node.left and not node.right and curSum == targetSum:
                res.append(stack.copy())
            dfs(node.left)
            dfs(node.right)
            # Remove current node after traversing, so we can backtrack
            curSum -= node.val
            stack.pop()
            
        dfs(root)
        return res

Time Complexity: $O(n)$, n is the number of nodes.
Space Complexity: $O(n)$, we store all nodes into stack in the worst case.