* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public:intpathSum(TreeNode* root,intsum) { unordered_map<int,int> count;//<curSum, num>count.insert({0,1});returndfs(root,0,sum,count); }intdfs(TreeNode *root,intcurSum,inttarget, unordered_map<...
res +=pathSum(root->right, sum, tmp); }returnres; }intpathSum(TreeNode* root,intsum){ unordered_map<longlong,longlong> mp; mp[sum] =0;returnpathSum(root, sum, mp); } };
public class Solution private int _sum; private int _count; public int PathSum(TreeNode root, int sum) _count = 0; _sum = sum; Travel(root, new List<int>()); return _count; private void Travel(TreeNode current, List<int> ret) if(current == null) return ; if(current.val == _...
递归左右节点时,这时候的sum值应该是原sum值减去当前节点的值。 class Solution: def hasPathSum(self, root, sum): """ :type root: TreeNode :type sum: int :rtype: bool """ if root is None: return False if (root.val == sum) and (root.left is None) and (root.right is None): re...
LeetCode Top 100 Liked Questions 437. Path Sum III (Java版; Easy) 题目描述 You are given a binary tree in which each node contains an integer value. Find the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must ...
leetcode 112. Path Sum 、 113. Path Sum II 、437. Path Sum III,112.PathSum自己的一个错误写法:只有左右节点都为NULL时才是叶子节点,所以这个代码在例子[1,2],1的右节点时就判断错误了,这个右节点虽然sum满足条件,但他本身不是叶子节点正确写法:113.PathSumII第
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...
1. Description Path Sum III 2. Solution Recursive /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
Solution: 首先我们从根节点出发,去遍历这个根节点下所有满足条件的路径数目,每次到达一个节点,将传入的参数sum减去root的val,这是一个递归。然后我们需要遍历所有的根节点,那么又需要一个递归去遍历。两层递归。 /** * Definition for a binary tree node. ...
publicclassSolution { privateint_max =int.MinValue; publicintMaxPathSum(TreeNode root) { MaxPath(root); return_max; } privateintMaxPath(TreeNode current) { if(current == null) return0; // 如果子树路径和为负则应当置0表示最大路径不包含子树 ...