113路径总和II

Posted 2020-11-24 xxswkl

题目: 给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

来源: https://leetcode-cn.com/problems/path-sum-ii/

法一: 自己的代码, 没有官方题解

思路: 递归实现,类似前序遍历

# 执行用时 :44 ms, 在所有 python3 提交中击败了99.18% 的用户
# 内存消耗 :14.2 MB, 在所有 python3 提交中击败了98.55%的用户
# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
from typing import List
class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        results = []
        res= []
        s = 0
        def recursion(root,s):
            # 如果遍历到一个分支的底了,则为空节点,返回True,继续遍历
            if root == None:
                return True
            res.append(root.val)
            # 注意不传s不行,全局变量s指向不可变对象,只能在里面调用,不能修改
            s = s + root.val
            a = recursion(root.left,s)
            b = recursion(root.right,s)
            # 如果到叶节点了,且和也满足条件,则存储
            if a and b:
                if s == sum:
                    results.append(res[:])
            s = s - root.val
            # 注意这里res是个全局变量,且因为指向可变对象所以可以修改,每次函数结束时,res中的
            # 内容不会自定返回到调用前的状态,所以需要出栈来恢复状态
            res.pop()
        recursion(root,s)
        return results
if __name__ == ‘__main__‘:
    duixiang = Solution()

    root = TreeNode(8)
    a = TreeNode(4)
    b = TreeNode(9)
    root.left = a
    root.right = b

    al = TreeNode(3)
    ar = TreeNode(6)
    a.left = al
    a.right = ar

    arl = TreeNode(5)
    arr = TreeNode(7)
    ar.left = arl
    ar.right = arr
    sum = 17
    a = duixiang.pathSum(root,sum)
    print(a)

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