The path sum is (3 + 5) + (3 + 1) = 12. Example 2: Input: [113, 221] Output: 4 Explanation: The tree that the list represents is: 3 \ 1 The path sum is (3 + 1) = 4. 还是二叉树的路径之和,但树的存储方式变了,使用一个三位的数字来存的,百位是该结点的深度,十位是该结点...
self.count+= (i%100)%10returnireturnnodedefIsLeaf(self,nums,node): dep= node/100pos= (node%100)/10next= (dep+1)*10 + pos*2next2= (dep+1)*10 + pos*2 -1foriinnums:ifi/10 == nextori/10 ==next2:returnFalsereturnTruedefpathSum(self, nums):""":type nums: List[int] :rtyp...
bool dfs(TreeNode *node, int sum, int curSum) { if (node == NULL) return false; if (node->left == NULL && node->right == NULL) return curSum + node->val == sum; return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val); } bo...
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]Path Sum Question 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. For example: Given the below binary tree and sum = 22,...
【CSON】LeetCode讲解 64. Minimum Path Sum发布于 2022-01-13 11:41 · 274 次播放 赞同添加评论 分享收藏喜欢 举报 力扣(LeetCode)算法编程程序员面试移动社交 写下你的评论... 还没有评论,发表第一个评论吧相关推荐 14:31 不是这假期也没告诉我是这么过的呀,早知道是这么个...
题目链接:Path Sum II | LeetCode OJ 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 and sum = 22, 5 /\ 48 //\ 11134 /\/\ ...
minSum=(~(unsignedint))>>; my_grid=grid; rowMax=grid.size(); colMax=grid[].size(); tra(,,); returnminSum; } }; 解法2:DP(还是不熟练,不太熟练递推dp和递归dp的区别,参考文章) dp[100][100];该dp数组记录的是每个位置上的最优解,即到达这一点的路径最小值。假设我们要求以grid[i][j...
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 ...
http://www.programcreek.com/2013/02/leetcode-binary-tree-maximum-path-sum-java/ 另外,这道题目的BST条件,似乎没什么用。因为如果全是负数,BST也没帮助了。 ** 总结: pre-order -> top-down post-order -> bottom-up in-order -> ? level-order -> bfs ...