] 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,...
Minimum Path Sum Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. SOLUTION 1: 相当基础的DP题目: This is a simpl...
right=None): self.val = val self.left = left self.right = right'''class Solution: def helper(self,root,acc,target,hashmap): if not root: return 0 count=0 acc+=root.val if acc==target: count+=1 if acc-target in hashmap: count+=hashmap[acc-...
class Solution { public int pathSum(TreeNode root, int targetSum) { if (root == null) return 0; return dfs(root, targetSum) + pathSum(root.left, targetSum) + pathSum(root.right, targetSum); } public int dfs(TreeNode root, long target) { if (root == null) return 0; int cnt...
class Solution: def pathSum(self, root, sum): """ :type root: TreeNode :type sum: int :rtype: List[List[int]] """ self.result = [] self.find_path(root, sum, []) return self.result def find_path(self, node, sum, tmp): if node is None: return if (node.val == sum) ...
class Solution(object): def dfs(self, root, mysum, target_sum): if not root:#当root只有一个儿子的时候 return False elif root.left == None and root.right == None: mysum += root.val#这里要把当前root节点的val加上之后再判断 if mysum == target_sum: ...
classSolution{privateMap<Long,Integer> map;privateintresult=, target =;publicintpathSum(TreeNode root,int targetSum){if(root ==null)return; target = targetSum; map =newHashMap(); map.put(0L,1); dfs(root, root.val);return result;}publicvoiddfs(TreeNode node,long value){ ...
right=None): # self.val = val # self.left = left # self.right = right class Solution: def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool: # 空结点不存在路径,直接返回 false if not root: return False # 如果是叶子结点,则仅当值等于 targetSum 时,才满足题意 if...
class Solution { public: bool hasPathSum(TreeNode *root, int sum) { if(root == nullptr){ return false; } sum -= root->val; if(root->left == nullptr && root->right == nullptr && sum ==0){ return true; } return hasPathSum(root->left,sum) || hasPathSum(root->right,sum);...
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...