Construction and Conversion

1. Construct Binary Tree from Preorder and Inorder Traversal (LeetCode #105)

1. Problem Description: The problem "Construct Binary Tree from Preorder and Inorder Traversal" on LeetCode #105 requires us to build a binary tree given its preorder and inorder traversal sequences.

Given two integer arrays, preorder and inorder, which represent the preorder and inorder traversal sequences of a binary tree, our task is to construct and return the binary tree.

For example: Input: preorder = [3,9,20,15,7] inorder = [9,3,15,20,7]

Output: 3 / 9 20 / 15 7

1. Optimized Java code for the problem "Construct Binary Tree from Preorder and Inorder Traversal":

import java.util.HashMap;
import java.util.Map;

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
        val = x;
    }
}

public class ConstructBinaryTreeFromPreorderAndInorder {
    // To improve the lookup time for elements in the inorder array
    Map<Integer, Integer> inorderMap;

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        // Construct a map to store the indexes of elements in the inorder array
        inorderMap = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            inorderMap.put(inorder[i], i);
        }
        return constructTree(preorder, 0, preorder.length - 1, 0);
    }

    private TreeNode constructTree(int[] preorder, int start, int end, int preStart) {
        // Base case
        if (start > end)
            return null;

        // Create a new node with the current element from preorder array
        TreeNode node = new TreeNode(preorder[preStart]);

        // Find the index of the current element in the inorder array
        int inIndex = inorderMap.get(node.val);

        // Calculate the number of elements in the left subtree
        int leftTreeSize = inIndex - start;

        // Build the left subtree recursively
        node.left = constructTree(preorder, start, inIndex - 1, preStart + 1);

        // Build the right subtree recursively
        node.right = constructTree(preorder, inIndex + 1, end, preStart + leftTreeSize + 1);

        return node;
    }
}

1. Detailed Description of the Algorithm and Interview Process Considerations:

The problem can be solved using a recursive approach by utilizing the preorder and inorder properties of the binary tree. Here are the steps to approach the problem:

• First, we should create a map to store the indexes of elements in the inorder array. This will allow us to find the index of any element in constant time while constructing the tree.

• Then, we define a recursive helper function constructTree that takes in the preorder array, the start and end indices of the current subtree in the inorder array, and the index of the current element in the preorder array.

• Inside the constructTree function, we handle the base case where the start index is greater than the end index. In this case, we return null as there are no elements to construct the subtree.

• Otherwise, we create a new node with the current element from the preorder array.

• Then, we find the index of the current element in the inorder array using the map created earlier.

• We calculate the number of elements in the left subtree by subtracting the start index from the inIndex.

• Next, we recursively build the left subtree by calling the constructTree function for the left subtree with updated indices.

• Similarly, we recursively build the right subtree by calling the constructTree function for the right subtree with updated indices.

• Finally, we return the node.

During the interview process, it's important to discuss and analyze the problem requirements, constraints, and approaches with the interviewer. It's also recommended to discuss the time and space complexity of the proposed solution and potential optimizations.

Additionally, you can demonstrate your problem-solving skills by discussing various test cases and edge cases (e.g., empty arrays, arrays with repeated elements) to ensure the algorithm handles all scenarios correctly.

2. Construct Binary Tree from Inorder and Postorder Traversal (LeetCode #106)

1. Problem Description: The problem "Construct Binary Tree from Inorder and Postorder Traversal" states that given two integer arrays, inorder and postorder, representing the inorder and postorder traversals of a binary tree respectively, we need to construct the binary tree and return its root node.

LeetCode Link: https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

1. Optimal Java Code:

class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        // mapping element to its index in inorder array for O(1) lookup
        Map<Integer, Integer> inorderMap = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            inorderMap.put(inorder[i], i);
        }
        
        return buildTreeHelper(postorder, 0, postorder.length - 1, inorder, 0, inorder.length - 1, inorderMap);
    }
    
    private TreeNode buildTreeHelper(int[] postorder, int postStart, int postEnd, int[] inorder, int inStart, int inEnd, 
                                     Map<Integer, Integer> inorderMap) {
        if (postStart > postEnd || inStart > inEnd) {
            return null;
        }
        
        int rootVal = postorder[postEnd];
        TreeNode root = new TreeNode(rootVal);
        
        // Find the index of rootVal in inorder array
        int rootIndexInorder = inorderMap.get(rootVal);
        
        // Calculate the number of nodes in left subtree
        int leftSubtreeNodes = rootIndexInorder - inStart;
        
        // Build the left and right subtrees recursively
        root.left = buildTreeHelper(postorder, postStart, postStart + leftSubtreeNodes - 1,
                                    inorder, inStart, rootIndexInorder - 1, inorderMap);
        root.right = buildTreeHelper(postorder, postStart + leftSubtreeNodes, postEnd -1,
                                     inorder, rootIndexInorder + 1, inEnd, inorderMap);
        
        return root;
    }
}

1. Detailed Explanation:

• The problem asks us to construct a binary tree using inorder and postorder traversals. In inorder traversal, we visit the left subtree first, then the root, and then the right subtree. In postorder traversal, we visit the left subtree, then the right subtree, and finally the root. The last element in postorder traversal is always the root element.

• To solve this problem, we start by creating a hashmap to map each element of the inorder array to its index. This will allow us to perform O(1) lookup to find the index of the root element in the inorder array.

• We then define a helper recursive function "buildTreeHelper" which takes the postorder and inorder arrays along with their respective start and end indices, and the inorderMap.

• In the helper function, we first check if the indices are valid. If not, we return null.

• We then find the root element using the last element of the postorder array, "postorder[postEnd]". We create a new TreeNode using this root value.

• Next, we find the index of the root element in the inorder array using the inorderMap. This index will divide the inorder array into left and right subtrees.

• We calculate the number of nodes in the left subtree by subtracting the starting index of the inorder array from the rootIndexInorder.

• We then recursively build the left and right subtrees by calling the buildTreeHelper function. For the left subtree, we pass the appropriate start and end indices from both the postorder and inorder arrays. For the right subtree, we pass the appropriate start and end indices from the postorder array, and the rootIndexInorder + 1 and inEnd indices from the inorder array.

• Finally, we return the root node of the constructed binary tree.

• The time complexity of this solution is O(n), where n is the number of nodes in the binary tree, and the space complexity is O(n) as well, considering the space required for the hashmap and the recursive function call stack during the recursion.

3. Convert Sorted Array to Binary Search Tree (LeetCode #108)

1. Problem Description: The problem "Convert Sorted Array to Binary Search Tree" is a LeetCode problem (Problem #108) that asks us to construct a binary search tree from a sorted array of integers.

Link to the problem on LeetCode: Convert Sorted Array to Binary Search Tree

1. Optimized Java Code with Detailed Comments:

// Definition for a binary tree node.
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
    }
}

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        return buildBST(nums, 0, nums.length - 1);
    }

    // Recursive helper function to build BST from sorted array
    private TreeNode buildBST(int[] nums, int left, int right) {
        if (left > right) {
            return null;
        }

        // Find the middle element of the array and create a new TreeNode
        int mid = left + (right - left) / 2;
        TreeNode node = new TreeNode(nums[mid]);

        // Recursively build left and right subtrees
        node.left = buildBST(nums, left, mid - 1);
        node.right = buildBST(nums, mid + 1, right);

        return node;
    }
}

1. Detailed Description and Algorithm:

The problem requires us to construct a balanced binary search tree (BST) from a sorted array. The key to solving this problem efficiently lies in finding the middle element of the array and creating a new TreeNode with that element as the root. We can then recursively build the left and right subtrees using the remaining elements on each side of the middle element.

Algorithm:

1. Create a public method sortedArrayToBST that takes in the sorted array nums as input and returns the root of the resulting BST.

2. In the sortedArrayToBST method, call a private recursive helper method buildBST with the array nums and the indices of the first and last elements as input.

3. The buildBST method handles the actual construction of the BST. It takes in the array nums and the indices left and right that represent the subrange of the array we are currently working with.

4. Base Case: If left is greater than right, return null as there are no more elements to process.

5. Find the middle index of the current subrange using mid = left + (right - left) / 2. This will be the index of the element that we use to construct the current TreeNode.

6. Create a new TreeNode with the value at nums[mid].

7. Recursively call buildBST for the left subrange of the array from left to mid - 1 and set the result as the left child of the current TreeNode.

8. Recursively call buildBST for the right subrange of the array from mid + 1 to right and set the result as the right child of the current TreeNode.

9. Return the constructed TreeNode.

This algorithm ensures that the BST is balanced because at each step, we divide the remaining array into two equal-sized halves. The time complexity of this solution is O(n), where n is the number of elements in the input array.

4. Convert Sorted List to Binary Search Tree (LeetCode #109)

1. Problem Description: The problem "Convert Sorted List to Binary Search Tree" asks us to convert a singly linked list into a balanced binary search tree. The given linked list is sorted in ascending order.

   LeetCode Link: Convert Sorted List to Binary Search Tree
2. Optimized Java Code:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 *
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    // Global variable to iterate through the linked list
    private ListNode current;
    
    public TreeNode sortedListToBST(ListNode head) {
        int size = getSize(head);
        current = head;
        
        return convertToBST(0, size - 1);
    }
    
    private int getSize(ListNode head) {
        int count = 0;
        ListNode temp = head;
        while (temp != null) {
            count++;
            temp = temp.next;
        }
        return count;
    }
    
    private TreeNode convertToBST(int start, int end) {
        if (start > end)
            return null;
        
        int mid = start + (end - start) / 2;
        
        TreeNode left = convertToBST(start, mid - 1);
        
        TreeNode root = new TreeNode(current.val);
        root.left = left;
        current = current.next;
        
        TreeNode right = convertToBST(mid + 1, end);
        root.right = right;
        
        return root;
    }
}

1. Algorithm Explanation:

   • To convert a sorted linked list into a balanced binary search tree, we can follow a recursive approach.

   • Firstly, we need to find the size of the linked list, so that we know how many elements are there.

   • Then, we start by initializing the current pointer to the head of the linked list.

   • Now, we can define a recursive function convertToBST which takes in the start and end indices of the linked list.

   • In the convertToBST function, we calculate the middle index as (start + end) / 2 and create a new tree node with the value at the current pointer.

   • Then, we recursively construct the left subtree by calling convertToBST on the left half of the linked list (start, mid - 1).

   • After constructing the left subtree, we move the current pointer to the next element in the linked list.

   • Finally, we recursively construct the right subtree by calling convertToBST on the right half of the linked list (mid + 1, end).

   • The base case for the recursion is when the start index is greater than the end index, where we return null.

   • In the main sortedListToBST function, we calculate the size of the linked list and set the current pointer to the head of the list.

   • Finally, we call the convertToBST function with start index 0 and end index (size - 1) to get the root of the constructed binary search tree.

   This approach has a time complexity of O(n), where n is the number of elements in the linked list, as we iterate through the linked list once to find its size and construct the binary search tree. The space complexity is O(log n) for the recursive call stack, as the height of the binary search tree is log(n) in the average case.

5. Serialize and Deserialize Binary Tree (LeetCode #297)

1. Problem Description: The problem "Serialize and Deserialize Binary Tree" on LeetCode (#297) requires us to design an algorithm to serialize and deserialize a binary tree. Serializing a binary tree means converting the tree into a string, and deserializing means converting the string back into a binary tree. LeetCode Link: https://leetcode.com/problems/serialize-and-deserialize-binary-tree/

2. Java Code: Here is the optimized Java code for the problem "Serialize and Deserialize Binary Tree" with detailed comments:

import java.util.*;

public class Codec {

    // Serialize the given binary tree
    public String serialize(TreeNode root) {
        if (root == null) {
            return "null"; // Base case: return "null" for null node
        }
        // Preorder traversal to serialize the tree
        return root.val + "," + serialize(root.left) + "," + serialize(root.right);
    }

    // Deserialize the given string to a binary tree
    public TreeNode deserialize(String data) {
        Queue<String> queue = new LinkedList<>(Arrays.asList(data.split(",")));
        return buildTree(queue);
    }

    // Helper function to build the binary tree from the serialized string
    private TreeNode buildTree(Queue<String> queue) {
        String val = queue.poll(); // Polling the next value from the queue
        if (val.equals("null")) {
            return null; // Base case: return null for "null"
        }
        TreeNode node = new TreeNode(Integer.parseInt(val)); // Create a new TreeNode with the parsed integer value
        node.left = buildTree(queue); // Recursively build the left subtree
        node.right = buildTree(queue); // Recursively build the right subtree
        return node; // Return the root of the binary tree
    }
}

1. Algorithm and Interview Thought Process:

• To solve this problem efficiently during an interview, we can use a depth-first search (DFS) approach along with the preorder traversal to serialize and deserialize the binary tree. The preorder traversal helps in maintaining the tree structure while serializing and deserializing.

• Serializing the Binary Tree:

  • For serializing the binary tree, we will start with the root node. If the node is null, we will represent it as "null" and return. For a non-null node, we will convert its value to a string and append it to the serialized string. Then, we will recursively serialize the left and right subtrees by making recursive calls to the function. After serializing both subtrees, we will return the serialized string.

• Deserializing the Binary Tree:

  • To deserialize the binary tree, we need to split the string at commas to separate the individual node values. We can use a queue to store the node values from the serialized string.

  • We will start by dequeuing the next value from the queue. If the dequeued value is "null", we will return null. Otherwise, we will convert the value to an integer and create a new TreeNode with that value.

  • We will then recursively build the left and right subtrees by making recursive calls to the buildTree function. The buildTree function will dequeue the next value from the queue and build the corresponding subtree recursively.

  • Finally, we will return the root of the deserialized binary tree.

• During the interview process, it is important to discuss the time and space complexity of the solution. In this case, the time complexity is O(N) for both serialization and deserialization, where N is the number of nodes in the binary tree. The space complexity is O(N) as well, as the serialized string representation requires O(N) space, and the recursive calls also use additional stack space.

6. Convert Binary Search Tree to Sorted Doubly Linked List (LeetCode #426)

1. Problem Description:

The problem "Convert Binary Search Tree to Sorted Doubly Linked List" on LeetCode (#426) asks us to convert a given binary search tree into a sorted doubly linked list. The transformed linked list should maintain the BST order.

Problem Link: https://leetcode.com/problems/convert-binary-search-tree-to-sorted-doubly-linked-list/

1. Optimized Java Code:

Here's the optimized Java code solution for the problem "Convert Binary Search Tree to Sorted Doubly Linked List":

class Solution {
    // Definition for a Node.
    static class Node {
        public int val;
        public Node left;
        public Node right;

        public Node() {}

        public Node(int val) {
            this.val = val;
        }
    }

    // Global variables to hold reference of the head and tail of the doubly linked list
    Node head = null;
    Node tail = null;

    public Node treeToDoublyList(Node root) {
        if (root == null) return null;
        
        // Convert the BST to a sorted doubly linked list using inorder traversal
        inorder(root);
        
        // Adjust the head and tail pointers to form a circular doubly linked list
        head.left = tail;
        tail.right = head;
        
        return head;
    }

    private void inorder(Node node) {
        if (node == null) return;
        
        // Recursively traverse the left subtree
        inorder(node.left);
        
        // Process the current node
        if (head == null) {
            // For the first node encountered, assign it as both head and tail
            head = node;
        } else {
            // For subsequent nodes, adjust the links accordingly
            tail.right = node;
            node.left = tail;
        }
        tail = node;
        
        // Recursively traverse the right subtree
        inorder(node.right);
    }
}

1. Algorithm Explanation and Interview Tips:

The algorithm to convert a binary search tree (BST) to a sorted doubly linked list is fairly straightforward:

1. Perform an inorder traversal of the BST.

2. During the traversal, maintain references to the head and tail of the resulting doubly linked list.

3. For each node encountered during the inorder traversal, adjust the links to connect it to the previous and next nodes in order.

4. Finally, adjust the head and tail pointers to form a circular doubly linked list.

During an interview, it is important to discuss and understand the problem requirements before jumping into coding. Clarify any doubts with the interviewer and ask for any additional constraints or edge cases. Draw the BST structure on paper or a whiteboard and discuss your approach aloud to demonstrate your thought process.

Explain the chosen algorithm thoroughly, mentioning its time and space complexity. Provide a dry run example to illustrate how the algorithm works. Test your code on a few test cases to verify its correctness.

Remember to write clean and well-commented code to make it easier for the interviewer to follow your thought process.