The Problem Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer or transmitted across a network connection link to be reconstructed later in the same or another computer environment. Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure. ...

The Problem A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root. The path sum of a path is the sum of the node’s values in the path. Given the root of a binary tree, return the maximum path sum of any non-empty path. ...

The Problem 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. Examples Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] Output: [3,9,20,null,null,15,7] 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. Depth First Search Solution func buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? { if preorder.isEmpty || inorder.isEmpty { return nil } let rootVal = preorder[0] let root = TreeNode(rootVal) guard let mid = inorder.firstIndex(of: rootVal) else { return nil } if mid > 0 { root.left = buildTree(Array(preorder[1...mid]), Array(inorder[0..<mid])) } if mid < inorder.count - 1 { root.right = buildTree(Array(preorder[mid + 1..<preorder.count]), Array(inorder[mid + 1..<inorder.count])) } return root } Explanation We can solve this problem by understanding how preorder and inorder traversal work. ...

The Problem Given the root of a binary search tree and an integer k, return the kth smallest value (1-indexed) among all the values of the nodes in the tree. Examples Input: root = [3,1,4,null,2], k = 1 Output: 1 Input: root = [5,3,6,2,4,null,null,1], k = 3 Output: 3 Constraints The number of nodes in the tree is n. 1 <= k <= n <= 10⁴ 0 <= Node.val <= 10⁴ Follow-up: If the BST is modified often (i.e., we can perform insert and delete operations) and you need to find the kth smallest element frequently, how would you optimize? ...

The Problem Given the root of a binary tree, determine if it is a valid binary search tree (BST). A valid BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node’s key. The right subtree of a node contains only nodes with keys greater than the node’s key. Both the left and right subtrees must also be binary search trees. A subtree of treeName is a tree consisting of a node in treeName and all of its descendants. ...