Leetcode 112. Path Sum 112. Path Sum1.有点回溯的感觉,试完左树试右树。 2.注意叶子节点的判断条件 3.root=NULL,sum=0时也要返回false。class HasPathSum { public: bool hasPathSum(TreeNode* root, int sum) { if (root == NULL) { return false; } pathSum = 0; return hasPathSumCore...
之后在leetcode上看到更简单的做法(-_-||) publicbooleanhasPathSum2(TreeNode root,intsum) {if(root ==null)returnfalse;if(root.left ==null&& root.right ==null&& sum - root.val == 0)returntrue;returnhasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum -root.val); ...
*/classSolution{public:boolhasPathSum(TreeNode* root,inttarget,int& cur){ cur += root->val;if(!root->left && !root->right){if(cur == target){returntrue; } }else{if(root->left &&hasPathSum(root->left, target, cur)){returntrue; }if(root->right &&hasPathSum(root->right, targe...
bool hasPathSum(TreeNode* root, int sum) { if(root == nullptr) return false; if(root->left == nullptr && root->right == nullptr && root->val == sum) return true; return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val); } 1. 2. 3...
Leetcode 112: Path Sum 会微积分的喵 西安电子科技大学 信息与通信工程硕士 来自专栏 · 每天学点算法 难度:Easy 相似题目: 437. Path Sum III 题目描述 给你一棵二叉树和一个整数sum,判断这棵树是否存在根节点到叶子节点,以至于路径上面的数和为sum。
Left, remainSum) { return true } // 当右子结点存在,且存在一条右子结点到叶子路径上所有值到和为 remainSum ,则满足题意 if root.Right != nil && hasPathSum(root.Right, remainSum) { return true } // 左右子结点都不满足题意,返回 false return false } 题目链接: Path Sum: leetcode.com...
LeetCode_112. Path Sum 112. Path Sum Easy Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. Note: A leaf is a node with no children....
returnresult+[path+[root.val]] else: returnself._pathSum(root.left,sum-root.val,path+[root.val],result)+self._pathSum(root.right,sum-root.val,path+[root.val],result) Leetcode 笔记系列的Python代码共享在https://github.com/wizcabbit/leetcode.solution...
今天介绍的是LeetCode算法题中Easy级别的第28题(顺位题号是112)。给定二叉树和整数sum,确定树是否具有根到叶路径,使得沿路径的所有值相加等于给定的sum。叶子节点是没有子节点的节点。例如: 给定以下二叉树和sum = 22, 5 / \ 4 8 / / \ 11 13 4 ...
easy版112,每次recursion少一个path node list的参数即可 【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:in...