SOLUTION 1: 使用递归解决,先把下一个可能要加的节点加入到path中,再使用递归依次计算即可。 View Code SOLUTION 2: 使用递归解决,如果只考虑加入当前节点,会更加简单易理解。递归的base case就是: 1. 当null的时候返回。 2. 当前节点是叶子 并且sum与root的值相同,则增加一个可能的解。 3. 如果没有解,将s...
求路径的和与某一特定的值时候对等即可,简单的dfs,代码如下: 1classSolution {2public:3vector<vector<int>> pathSum(TreeNode* root,intsum) {4vector<int>res;5dfs(root, res, sum);6returnret;7}89voiddfs(TreeNode * root, vector<int> res,intleft)10{11if(!root)return;12if(!root->left &&...
Can you solve this real interview question? Path Sum II - 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
解法:这道题和Path Sum很相似,所不同的是,我们需要找到所有的路径并且记录下这些路径,但是方法还是一样的;对于一颗二叉树T,找一条从根节点到叶子节点的路径满足和为sum,即递归的找根节点的左右子树中是否有一条从左右节点到根节点的路径满足节点值得和为sum-T.val; Java AI检测代码解析 /** *...
LeetCode: 113. Path Sum II 题目描述 Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum. For example: Given the below binary tree and sum = 22, AI检测代码解析 5 / \ 4 8 ...
// possible "right" solution. } return list; } } Approach1:Recursive solution we use a helper DFS to get our solution recursively. class Solution { public List<List<Integer>> pathSum(TreeNode root, int sum) { if(root == null) return new ArrayList(); ...
【Solution】 # Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = NoneclassSolution:def__init__(self):self.result=[]defpathSum(self,root:TreeNode,sum:int)->bool:# DFS & recursion |defrecursion(root,path...
Link:https://leetcode.com/problems/path-sum-ii/ Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum. For example: Given the below binary tree and sum = 22, 5 / \48 / / \11134 ...
[Leetcode][python]Path Sum II/路径总和 II 题目大意 将根到叶子的路径和为sum的路径都枚举出来。 解题思路 递归,并且用了python函数嵌套,有关函数嵌套可以看这一篇文章 其实一开始不想项标准答案一样用函数嵌套,毕竟别的语言可能不支持,以后看答案不方便,但是如果把list_all放在全局,需要每轮都去清空它,而...
测试用例: https://leetcode.com/problems/path-sum-ii/description/ """ # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def pathSum(self, root, sum): """...