Leetcode

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] ...

94. Binary Tree Inorder Traversal Given the root of a binary tree, return the inorder traversal of its nodes’ values. Example 1: Input: root = [1,null,2,3] Output: [1,3,2] 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: [1,2] ...

144. Binary Tree Preorder Traversal Given the root of a binary tree, return the preorder traversal of its nodes’ values. Example 1: Input: root = [1,null,2,3] Output: [1,2,3] Example 2: Input: root = [] Output: [] Example 3: Input: root = [1] Output: [1] Example 4: Input: root = [1,2] Output: [1,2] Example 5: Input: root = [1,null,2] Output: [1,2] ...

149. Max Points on a Line Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane, return the maximum number of points that lie on the same straight line. Example 1: Input: points = [[1,1],[2,2],[3,3]] Output: 3 Example 2: Input: points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]] Output: 4 Constraints: 1 <= points.length <= 300 points[i].length == 2 -104 <= xi, yi <= 104 All the points are unique. /** * 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 TreeNode sortedArrayToBST(int[] nums) { if (nums.length == 0) return null; return buildBST(nums, 0, nums.length - 1); } TreeNode buildBST(int[] nums, int lo, int hi) { if (hi < lo) { return null; } TreeNode node = new TreeNode(); int mid = lo + (hi - lo) / 2; node.val = nums[mid]; node.left = buildBST(nums, lo, mid - 1); node.right = buildBST(nums, mid + 1, hi); return node; } }

108. Convert Sorted Array to Binary Search Tree Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree. A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one. Example 1: Input: nums = [-10,-3,0,5,9] Output: [0,-3,9,-10,null,5] Explanation: [0,-10,5,null,-3,null,9] is also accepted: ...

110. Balanced Binary Tree Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the left and right subtrees of every node differ in height by no more than 1. Example 1: Input: root = [3,9,20,null,null,15,7] Output: true Example 2: Input: root = [1,2,2,3,3,null,null,4,4] Output: false Example 3: Input: root = [] Output: true Constraints: ...

111. Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. Note: A leaf is a node with no children. Example 1: Input: root = [3,9,20,null,null,15,7] Output: 2 Example 2: Input: root = [2,null,3,null,4,null,5,null,6] Output: 5 Constraints: The number of nodes in the tree is in the range [0, 105]. -1000 <= Node.val <= 1000 /** * 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 { int min = Integer.MAX_VALUE; public int minDepth(TreeNode root) { if (root == null) return 0; findMinDepth(root, 1); return min; } void findMinDepth(TreeNode node, int depth) { if (node == null) return; if (node.left == null && node.right == null) { min = Math.min(depth, min); } findMinDepth(node.left, depth + 1); findMinDepth(node.right, depth + 1); } } Solution 2021-08-23 /** * 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 { int min = Integer.MAX_VALUE; public int minDepth(TreeNode root) { findMinDepth(root, 1); return min == Integer.MAX_VALUE ? 0: min; } void findMinDepth(TreeNode node, int level) { if (node == null) { return; } if (node.left == null && node.right == null) { min = Math.min(min, level); return; } findMinDepth(node.left, level + 1); findMinDepth(node.right, level + 1); } } Solution with Stack 2021-11-11 /** * 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 int minDepth(TreeNode root) { Stack<Pair<TreeNode, Integer>> stack = new Stack<>(); if (root == null) { return 0; } stack.push(new Pair<>(root, 1)); int min = Integer.MAX_VALUE; while (stack.size() > 0) { Pair<TreeNode, Integer> pair = stack.pop(); TreeNode node = pair.getKey(); Integer currentMin = pair.getValue(); if (node.left == null && node.right == null) { min = Math.min(min, currentMin); continue; } if (node.left != null) { stack.push(new Pair<>(node.left, currentMin + 1)); } if (node.right != null) { stack.push(new Pair<>(node.right, currentMin + 1)); } } return min; } }

![https://leetcode.com/problems/same-tree/] Given the roots of two binary trees p and q, write a function to check if they are the same or not. Two binary trees are considered the same if they are structurally identical, and the nodes have the same value. Example 1: Input: p = [1,2,3], q = [1,2,3] Output: true Example 2: Input: p = [1,2], q = [1,null,2] Output: false Example 3: Input: p = [1,2,1], q = [1,1,2] Output: false ...