Tree Traversal

11. Binary Tree Paths (257)

1. The problem "Binary Tree Paths" can be found on LeetCode at the following link: https://leetcode.com/problems/binary-tree-paths/

2. Here is the optimized Java code for the problem "Binary Tree Paths (257)" with detailed comments:

import java.util.ArrayList;
import java.util.List;

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

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

public class BinaryTreePaths {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> paths = new ArrayList<>();
        
        if (root == null) {
            return paths;  // Return empty list if root is null
        }
        
        dfs(root, "", paths);  // Start the depth-first search (dfs) traversal
        
        return paths;
    }
    
    private void dfs(TreeNode node, String path, List<String> paths) {
        if (node.left == null && node.right == null) {
            paths.add(path + node.val);  // If node is a leaf node, add the path to the result list
        }
        
        if (node.left != null) {
            dfs(node.left, path + node.val + "->", paths);  // Recursively process the left child
        }
        
        if (node.right != null) {
            dfs(node.right, path + node.val + "->", paths);  // Recursively process the right child
        }
    }
}

1. Detailed description of the algorithm and how to think through it during an interview process:

   • The problem is asking for all root-to-leaf paths in a binary tree.

   • One approach to solving this problem is to use depth-first search (dfs) on the binary tree.

   • During the dfs traversal, we keep track of the current path from the root to the current node.

   • When we reach a leaf node, we add the current path to the result list.

   • To implement this, we can use a helper function that takes in the current node, the current path, and the result list as parameters.

   • The helper function checks if the current node is a leaf node. If it is, it adds the current path to the result list.

   • If the current node has a left child, we recursively call the helper function with the left child, the updated path (including the value of the current node and an arrow "->"), and the result list.

   • If the current node has a right child, we recursively call the helper function with the right child, the updated path, and the result list.

   • Finally, we start the dfs traversal from the root node by calling the helper function.

   • The result list will contain all the root-to-leaf paths in the binary tree.

   • This approach has a time complexity of O(n), where n is the number of nodes in the binary tree, since we visit each node exactly once during the dfs traversal.

   • During an interview, it's important to communicate the approach clearly and ensure that the code is correctly implemented with proper variable handling and base case checks. Additionally, providing a clear explanation of the time complexity helps to demonstrate understanding of algorithm analysis.

12. Lowest Common Ancestor of a Binary Tree (236)

1. Problem Description: The problem "Lowest Common Ancestor of a Binary Tree" asks us to find the lowest common ancestor (LCA) of two given nodes in a binary tree. The LCA is the lowest node in the tree that has both nodes as descendants.

LeetCode Problem Link: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-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 lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // Base case: if root is null or root matches either p or q, return root
        if (root == null || root == p || root == q) {
            return root;
        }
        
        // Recursively find the LCA in the left and right subtrees
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        
        // If both left and right are not null, it means p and q are found in different subtrees
        // So, the current root is the LCA
        if (left != null && right != null) {
            return root;
        }
        
        // If only one of left or right is not null, it means found either p or q (or both) in one subtree
        // So, the subtree's result is the LCA
        if (left != null) {
            return left;
        }
        if (right != null) {
            return right;
        }
        
        // If none of the above conditions are true, it means neither p nor q is found in the subtree
        // Return null to indicate no LCA found
        return null;
    }
}

1. Algorithm and Thought Process:

To solve this problem, we can use a recursive approach known as the "Divide and Conquer" technique.

The idea is to traverse the tree from the root to its children recursively until we find either p or q. If either p or q is found, we return it to its parent. Then, if the parent receives both nodes from its left and right subtrees, it means the parent is the lowest common ancestor (LCA).

The algorithm can be summarized as follows:

• If the current node is null or matches either p or q, return the current node.

• Recursively find the LCA in the left subtree and store the result in a variable called left.

• Recursively find the LCA in the right subtree and store the result in a variable called right.

• If both left and right are not null, it means p and q are found in different subtrees, so the current root is the LCA.

• If only one of left or right is not null, it means either p or q (or both) is found in one subtree, so the subtree's result is the LCA.

• If none of the above conditions are true, it means neither p nor q is found in the subtree, so return null to indicate no LCA found.

During an interview, it is essential to communicate your thought process and approach to the interviewer. You can explain the divide and conquer technique as the underlying principle for the solution. Additionally, you can discuss the time complexity of the solution (which is O(N), as we have to visit each node once) and any potential optimizations.

13. Maximum Depth of Binary Tree (104)

1. Problem Description: The problem "Maximum Depth of Binary Tree" (104) on Leetcode asks us to find the maximum depth (height) of a binary tree. The maximum depth of a binary tree is defined as the length of the longest path from the root node to any leaf node.

Link to the problem: https://leetcode.com/problems/maximum-depth-of-binary-tree/

1. Optimized Java Code:

/**
 * 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 maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        int leftDepth = maxDepth(root.left);
        int rightDepth = maxDepth(root.right);
        
        // Return the maximum depth between left and right subtrees, plus the root node
        return Math.max(leftDepth, rightDepth) + 1;
    }
}

1. Detailed Description: The problem is solved using a recursive approach. We make use of the concept of maximum depth of a binary tree, which is defined as the length of the longest path from the root node to any leaf node.

In the recursive solution, we check if the root node is null, which serves as the base case for our recursion. If the root node is null, it means there are no nodes in the binary tree, and we return 0 as the maximum depth.

If the root node is not null, we recursively calculate the maximum depth of the left and right subtrees. By making recursive calls to the maxDepth() function with the left and right child nodes as arguments, we effectively traverse down the binary tree, calculating the maximum depth for each subtree.

Finally, we return the maximum depth between the left and right subtrees, plus 1 to account for the current root node. This is done using the Math.max() function. This recursive process continues until we reach the leaf nodes, where the base case is triggered and 0 is returned for the depth of the empty subtree.

During interviews, it's important to discuss the problem with the interviewer, understand the problem statement clearly, and propose a solution. When tackling this problem, you can start by discussing the problem requirements, clarifying any doubts, and explaining your approach. Then you can proceed to implement the solution using the recursive method mentioned above.

Remember to test your implementation with sample inputs and edge cases to ensure it works correctly.

14. Validate Binary Search Tree (98)

Sure! Here are the details you requested:

1. Problem Description and LeetCode Link: Validate Binary Search Tree (98): 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.

LeetCode Link: https://leetcode.com/problems/validate-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 boolean isValidBST(TreeNode root) {
        return isValidBST(root, null, null);
    }
    
    private boolean isValidBST(TreeNode node, Integer min, Integer max) {
        // Base case: empty tree is a valid BST
        if (node == null) {
            return true;
        }
        
        // Check if current node violates the BST property
        if ((min != null && node.val <= min) || (max != null && node.val >= max)) {
            return false;
        }
        
        // Recursively check the left and right subtrees
        return isValidBST(node.left, min, node.val) && isValidBST(node.right, node.val, max);
    }
}

1. Algorithm and Explanation:

To determine if a binary tree is a valid binary search tree (BST), we can employ a recursive approach.

The main idea is to traverse the tree, keeping track of the valid range of values that each node's value must fall within. For any given node, its value must be within the range defined by the maximum value in its left subtree and the minimum value in its right subtree.

We start with the initial range of minimum and maximum as null, indicating that there are no constraints on the first node. Then, as we traverse the tree, we update the range based on the current node's value and its position in the tree.

If at any point during the traversal, we encounter a node whose value violates the BST property (i.e., it is less than or equal to the minimum or greater than or equal to the maximum), we can conclude that the tree is not a valid BST.

The isValidBST() function is a helper function that calls the recursive helper function isValidBST() passing the root node and the initial ranges. This helper function checks if a subtree rooted at a given node is a valid BST based on the specified range constraints.

In terms of time complexity, the algorithm traverses each node exactly once, so the time complexity is O(n), where n is the number of nodes in the tree. The space complexity is O(n) as well, due to the recursive stack space used during the traversal.

During an interview, it's essential to understand and explain the algorithm as mentioned above. Additionally, consider discussing edge cases like an empty tree, trees with identical node values, or nodes with INT_MIN and INT_MAX values to ensure the solution handles all possible scenarios.

15. Symmetric Tree (101)

1. Problem Description: The problem "Symmetric Tree" (101) on LeetCode involves determining whether a binary tree is symmetric. A binary tree is considered symmetric if its left and right subtrees are mirror images of each other.

LeetCode link: https://leetcode.com/problems/symmetric-tree/

1. Optimized Java Code: Here is the optimized Java code for the "Symmetric Tree" problem:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        
        return isMirror(root.left, root.right);
    }
    
    private boolean isMirror(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null) {
            return true;
        }
        if (t1 == null || t2 == null) {
            return false;
        }
        
        return (t1.val == t2.val) &&
            isMirror(t1.left, t2.right) &&
            isMirror(t1.right, t2.left);
    }
}

1. Description of the Algorithm and Interview Process Approach: This problem can be approached using a recursive algorithm. The recursive function isMirror checks if two given tree nodes and their respective subtrees are mirror images of each other.

The algorithm follows these steps:

• If the given root node is null, it means the tree is symmetric, so we return true.

• Otherwise, we call the isMirror function recursively with the left subtree and right subtree of the root.

• In the isMirror function:

  • If both nodes are null, it means the subtrees are symmetric, so we return true.

  • If one of the nodes is null but the other is not, it means the subtrees are not symmetric, so we return false.

  • If both nodes have different values, it means the subtrees are not symmetric, so we return false.

  • Otherwise, we recursively check if the left subtree of the first node is mirror to the right subtree of the second node, and vice versa.

During an interview, it is essential to explain the recursive nature of the algorithm and how it checks for symmetry. Additionally, discuss time complexity (O(n)) and space complexity (O(log n) on average, O(n) in worst case) and how the code handles edge cases, such as an empty tree (where symmetry is assumed). Providing a clear and concise explanation of the algorithm and code organization will demonstrate problem-solving skills and clarity of thought.