int val) { if (root == nullptr) { return 0; } val = (val << 1) | root->val; if (root->left == nullptr && root->right == nullptr) { return val; } return dfs(root->left, val) + dfs(root->right, val); } int sumRootToLeaf(TreeNode* root) { return dfs(root, 0);...
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/...
* Source : https://oj.leetcode.com/problems/sum-root-to-leaf-numbers/ * * * Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. * An example is the root-to-leaf path 1->2->3 which represents the number 123. * * Find the tota...
Given therootof a binary tree, consider allroot to leaf paths: paths from the root to any leaf. (A leaf is a node with no children.) Anodeisinsufficientif every such root to leaf path intersecting thisnodehas sum strictly less thanlimit. Delete all insufficient nodes simultaneously, and re...
Can you solve this real interview question? Sum Root to Leaf Numbers - You are given the root of a binary tree containing digits from 0 to 9 only. Each root-to-leaf path in the tree represents a number. * For example, the root-to-leaf path 1 -> 2 -> 3
Can you solve this real interview question? Sum Root to Leaf Numbers - You are given the root of a binary tree containing digits from 0 to 9 only. Each root-to-leaf path in the tree represents a number. * For example, the root-to-leaf path 1 -> 2 -> 3
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22. 【解答】没什么可说的,见代码: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 public class Solution { public boolean hasPathSum(TreeNode root, int sum) { if (null==root) return false; List<Integer>...
Sum Root to Leaf Numbers 2 4 tree DFS 130 Surrounded Regions 4 3 array BFS DFS 131 Palindrome Partitioning 3 4 string DFS 132 Palindrome Partitioning II 4 3 string DP 本文参与 腾讯云自媒体同步曝光计划,分享自作者个人站点/博客。 原始发表:2016-08-24 ,如有侵权请联系 cloudcommunity@tencent.com ...
0129 Sum Root to Leaf Numbers Go 58.6% Medium 0130 Surrounded Regions Go 35.8% Medium 0131 Palindrome Partitioning Go 62.3% Medium 0132 Palindrome Partitioning II 33.6% Hard 0133 Clone Graph 50.5% Medium 0134 Gas Station 45.0% Medium 0135 Candy Go 40.7% Hard 0136 Single Number Go ...
1017 Convert to Base -2 Go 59.0% Medium 1018 Binary Prefix Divisible By 5 47.7% Easy 1019 Next Greater Node In Linked List Go 57.4% Medium 1020 Number of Enclaves Go 57.7% Medium 1021 Remove Outermost Parentheses Go 77.9% Easy 1022 Sum of Root To Leaf Binary Numbers 67.2% Easy ...