import java.util.ArrayList; import java.util.Arrays; import java.util.List; /** * Source : https://oj.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 ...
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 andsum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] 解体思路:利用递归深度优先...
* };*/classSolution {public: vector<vector<int> > pathSum(TreeNode *root,intsum) { vector<vector<int> >result; vector<int>intermediate; dfs(root, sum, intermediate, result);returnresult; }voiddfs(TreeNode *root,intgap, vector<int> &intermediate, vector<vector<int> > &result){if(root...
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
if(root->left)helper(root->left,value,v); if(root->right)helper(root->right,value,v); v.pop_back(); } vector<vector<int>> pathSum(TreeNode* root,intsum) { vector<int> v; helper(root,sum,v); returnres; } }; 想法: 还需要pop_back(), 理解!
[5,4,11,2], [5,8,4,5] ] 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 题意:给定一颗二叉树,找出所有从根节点到叶子节点满足其上所有节点和为指定值的路径; 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...
本期例题:LeetCode 112 - Path Sum(Easy) 给定一个二叉树和一个目标和,判断该树中是否存在根结点到叶结点的路径,这条路径上所有结点值相加等于目标和。返回 true 或者 false。 示例: 给定如下二叉树,以及目标和 sum = 22。以下二叉树存在路径 5->4->11->2 满足目标和。
[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...
return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val); } bool hasPathSum(TreeNode *root, int sum) { return dfs(root, sum, 0); } }; 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. ...
与39.组合总和相同,终止条件为sum > target和sum == target。 代码如下: if (sum > target) { // 这个条件其实可以省略 return; } if (sum == target) { result.push_back(path); return; } sum > target这个条件其实可以省略,因为和在递归单层遍历的时候,会有剪枝的操作,下面会介绍到。