Given therootof a binary tree, consider all root to leaf paths: paths from the root to any leaf. (A leaf is a node with no children.) Anodeis insufficient if every such root to leaf path intersecting thisnodehas sum strictly less thanlimit. Delete all insufficient nodes simultaneously, an...
Github 同步地址: https://github.com/grandyang/leetcode/issues/1080 参考资料: https://leetcode.com/problems/insufficient-nodes-in-root-to-leaf-paths/ https://leetcode.com/problems/insufficient-nodes-in-root-to-leaf-paths/discuss/308326/JavaC%2B%2BPython-Easy-and-Concise-Recursion LeetCode All ...
The root-to-leaf path 1->3 represents the number 13. Return the sum = 12 + 13 = 25. 这道题就是一个简单的DFS深度优先遍历。 注意对叶子节点的判断。 代码如下: AI检测代码解析 import java.util.ArrayList; import java.util.List; /*class TreeNode { int val; TreeNode left; TreeNode right...
*/classSolution{public:voiddfs(int&result,intnum,TreeNode*node){//result表示全部路径的组合//num表示根到node的父节点的组合数if(node){num=10*num+node->val;//计算从根到当前节点的组合数if(node->left==NULL&&node->right==NULL){result+=num;//已经找到一个条路径的组合数,累加到result上return;...
Constraints: The total number of nodes is between [1, 5 * 10^4]. Each node has a unique value. 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/find-root-of-n-ary-tree 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
The root-to-leaf path 4->9->1 represents the number 491. The root-to-leaf path 4->0 represents the number 40. Therefore, sum = 495 + 491 + 40 = 1026. 【Idea】 跟求path的题差不多, 只不过这里是num*10+node.val 在每层递归里作参数传递罢辽。
(root,"",list);returnlist;}publicvoidpreOrder(TreeNode node,String path,List<String>list){if(node.left==null&&node.right==null){list.add(path+node.val);}if(node.left!=null){preOrder(node.left,path+node.val+"->",list);}if(node.right!=null){preOrder(node.right,path+node.val+"-...
The root-to-leaf path 4->0 represents the number 40. Therefore, sum = 495 + 491 + 40 = 1026. 利用递归得到从根节点到叶节点的所有路径,并计算路径之和。 classSolution{ public: intsumNumbers(TreeNode* root){ intsum=0; vector<int> path; ...
然后就是程序了,参考了LeetCode上的程序,然后写了个我觉得最好理解的贴在这里,顺便带上完整测试程序: #include<iostream> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} ...
Left, pathSum + root.Left.Val) } // 如果右子结点不为空,则递归处理右子结点 if root.Right != nil { result += dfs(root.Right, pathSum + root.Right.Val) } return result } 题目链接: Sum Root to Leaf Numbers : leetcode.com/problems/s 求根节点到叶节点数字之和: leetcode.cn/...