遍历过程中累加节点值,当到达任意叶节点的时候,进行判断。 [Code] 1:boolhasPathSum(TreeNode*root,intsum){ 2:returnhasPathSum(root,0,sum); 3:}4:boolhasPathSum(TreeNode*root,intsum,inttarget){5:if(root==NULL)returnfalse;6:sum+=root->val;7:if(root->left==NULL&&root->right==NULL)//...
classSolution { public: vector<vector<int> > pathSum(TreeNode *root,intsum) { // Start typing your C/C++ solution below // DO NOT write int main() function result.clear(); if(!root)returnresult; intcurSum = sum; vector<int> curVec; myPathSum(root,curSum,curVec); returnresult; ...
return hasPathSum(root,sum,value); } bool hasPathSum(TreeNode* root,int sum,int value){ if(root == NULL){ if(value == sum) return true; else return false; } bool left = hasPathSum(root->left,sum,value + root->val); bool right = hasPathSum(root->right,sum,value + root->v...
代码语言:c 代码运行次数:0 运行 AI代码解释 #Definitionfora binary tree node.#classTreeNode(object):#def__init__(self,x):#self.val=x#self.left=None#self.right=NoneclassSolution(object):defhasPathSum(self,root,sum):""":type root:TreeNode:type sum:int:rtype:bool"""ifnot root:returnFa...
class Solution { public: int minPathSum(vector<vector<int> > &grid) { int r = grid.size(); if (r < 1) return 0; int c = grid[0].size(); vector<int> table(c); if (c-1>=0) table[c-1] = grid[r-1][c-1];
classSolution {public:intminPathSum(vector<vector<int> > &grid) {intm=grid.size();if(m==0)return0;intn=grid[0].size(); vector<int>ivec(n); vector<vector<int>>f(m, ivec); f[0][0]=grid[0][0];for(intki=1;ki<m;ki++) ...
int minPathSum(vector<vector<int> > &grid) { // Start typing your C/C++ solution below // DO NOT write int main() function int n=grid.size(); if(n<=0) return 0; int m=grid[0].size(); vector<vector<long> >M; M.resize(n); for(int i=0;i<n;i++){ M[i].resize(m,...
class Solution:def minPathSum(self, grid: List[List[int]]) -> int: m, n = len(grid), len(grid[0])for i in range(1, m): grid[i][0] += grid[i - 1][0]for i in range(1, n): grid[0][i] += grid[0][i - 1]...
输入: path = “/a/./b/../.../c/” 输出: “/c” 解题思路 栈 参考: https://shenjie1993.gitbooks.io/leetcode-python/071%20Simplify%20Path.html...c' >>> a = ['','a','b'] >>> '/'.join(a) '/a/b 代码 class Solution(object): def simplifyPath(self, path...): ""...
// 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(); ...