LeetCode | 0113. Path Sum II sum path II [Python]

LeetCode 0113. Path Sum II Total Medium [II] [path] [back] Python

Problem

LeetCode

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

problem

Power button

Given a binary tree and a target and to find the sum of all paths from the root node of the leaf node is equal to a given target and path.

Description: leaf node is a node has no child nodes.

Example:
Given the following binary tree, and the target and sum = 22,

          5
         / \
        4   8
       /   / \
      11  13  4
     /  \    / \
    7    2  5   1

return:

[
   [5,4,11,2],
   [5,8,4,5]
]

Thinking

Backtracking

先序遍历二叉树，记录路径。
符合条件的加入 res 中。
回溯记得要删除当前节点。

Time complexity: O (n-), n-number of binary tree nodes.
Space complexity: O (n-), the worst case, a binary tree degenerate into a single linked list.

Python3 Code

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        res, path = [], []

        # 先序遍历：根左右
        def recur(root, target):
            if not root:
                return
            
            path.append(root.val)
            target -= root.val
            # 找到路径
            if target == 0 and not root.left and not root.right:
                res.append(list(path))  # 复制了一个 path 加入到 res 中，这样修改 path 不影响 res
            recur(root.left, target)
            recur(root.right, target)
            # 向上回溯，需要删除当前节点
            path.pop()
        
        recur(root, sum)
        return res

GitHub link

Python