235. Lowest Common Ancestor of a Binary Search Tree Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).” ...
114. Flatten Binary Tree to Linked List Given the root of a binary tree, flatten the tree into a “linked list”: The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null. The “linked list” should be in the same order as a pre-order traversal of the binary tree. Example 1: Input: root = [1,2,5,3,4,null,6] Output: [1,null,2,null,3,null,4,null,5,null,6] Example 2: Input: root = [] Output: [] Example 3: Input: root = [0] Output: [0] Constraints: ...
1441. Build an Array With Stack Operations Given an array target and an integer n. In each iteration, you will read a number from list = {1,2,3…, n}. Build the target array using the following operations: Push: Read a new element from the beginning list, and push it in the array. Pop: delete the last element of the array. If the target array is already built, stop reading more elements. Return the operations to build the target array. You are guaranteed that the answer is unique. ...
199. Binary Tree Right Side View Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. Example 1: Input: root = [1,2,3,null,5,null,4] Output: [1,3,4] Example 2: Input: root = [1,null,3] Output: [1,3] Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 100]. -100 <= Node.val <= 100 Recursive ...
107. Binary Tree Level Order Traversal II Given the root of a binary tree, return the bottom-up level order traversal of its nodes’ values. (i.e., from left to right, level by level from leaf to root). Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[15,7],[9,20],[3]] Example 2: Input: root = [1] Output: [[1]] Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 2000]. -1000 <= Node.val <= 1000 Jamboard ...
102. Binary Tree Level Order Traversal Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level). Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]] Example 2: Input: root = [1] Output: [[1]] Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 2000]. -1000 <= Node.val <= 1000 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<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> levels = new ArrayList<>(); levelOrder(root, levels, 0); return levels; } void levelOrder(TreeNode node, List<List<Integer>> paths, int k) { if (node == null) return; if (paths.size() == k) { paths.add(new ArrayList<>()); } paths.get(k).add(node.val); levelOrder(node.left, paths, k + 1); levelOrder(node.right, paths, k + 1); } }
743. Network Delay Time You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target. We will send a signal from a given node k. Return the time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1. ...
787. Cheapest Flights Within K Stops There are n cities connected by m flights. Each flight starts from city u and arrives at v with a price w. Now given all the cities and flights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to k stops. If there is no such route, output -1. Example 1: Input: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]] src = 0, dst = 2, k = 1 Output: 200 Explanation: The graph looks like this: ...
105. Construct Binary Tree from Preorder and Inorder Traversal Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree. Example 1: Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] Output: [3,9,20,null,null,15,7] Example 2: Input: preorder = [-1], inorder = [-1] Output: [-1] Constraints: 1 <= preorder.length <= 3000 inorder.length == preorder.length -3000 <= preorder[i], inorder[i] <= 3000 preorder and inorder consist of unique values. Each value of inorder also appears in preorder. preorder is guaranteed to be the preorder traversal of the tree. inorder is guaranteed to be the inorder traversal of the tree. Jamboard ...
145. Binary Tree Postorder Traversal Given the root of a binary tree, return the postorder traversal of its nodes’ values. Example 1: Input: root = [1,null,2,3] Output: [3,2,1] Example 2: Input: root = [] Output: [] Example 3: Input: root = [1] Output: [1] Example 4: Input: root = [1,2] Output: [2,1] Example 5: Input: root = [1,null,2] Output: [2,1] ...