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Given the root of a binary tree, return all root-to-leaf paths in any order. A leaf is a node with no children. Example 1: Input: root = [1,2,3,null,5] Output: ["1->2->5","1->3"] Example 2: Input: root = [1] Output: ["1"] Constraints: The number of nodes in the tree is in the range [1, 100]. -100 <= Node.val <= 100 Solution /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public List<String> binaryTreePaths(TreeNode root) { List<String> res = new ArrayList<>(); dfs(root, res, ""); return res; } void dfs(TreeNode node, List<String> res, String s) { if (node == null) { return; } if (s.length() == 0) { s += "" + node.val; } else { s += "->" + node.val; } if (node.left == null && node.right == null) { res.add(s); return; } dfs(node.left, res, s); dfs(node.right, res, s); } }

Given two sparse vectors, compute their dot product. Implement class SparseVector: SparseVector(nums) Initializes the object with the vector nums dotProduct(vec) Compute the dot product between the instance of SparseVector and vec A sparse vector is a vector that has mostly zero values, you should store the sparse vector efficiently and compute the dot product between two SparseVector. Follow up: What if only one of the vectors is sparse? ...