Preorder always tells you the next root. Inorder tells you where to split into left and right subtrees. Using left and right boundaries avoids creating new arrays and reduces the complexity from O(n²) to O(n).

A BST's inorder traversal visits nodes in ascending order. The kth node visited during inorder traversal is the kth smallest element. The recursive solution is concise and easy to understand. The iterative solution uses a stack to simulate recursion and is the standard interview approach.

Comparing a node with only its parent is not sufficient. Every node must satisfy constraints inherited from all its ancestors. Passing a valid (lower, upper) range during DFS elegantly enforces the BST rules. This is the standard interview solution and is both simple and optimal.

This is an excellent example of a DFS problem where the trick is not the traversal itself, but what information you carry along the path. By passing the maximum value seen so far, each node can determine independently whether it is "good," resulting in a simple and efficient O(n) solution.

Whenever a binary tree problem asks for something "per level", BFS should be one of your first thoughts. For the right side view: Traverse level by level. The last node of every level is visible from the right. Record it and continue.

This problem naturally asks us to visit nodes level by level, making BFS the ideal traversal. This solution runs in O(n) time with O(n) auxiliary space and is the standard interview solution for this problem.

Binary Search Trees (BSTs) are special because they allow us to make decisions without exploring every node. This problem is a great example of taking advantage of that property to find the Lowest Common Ancestor (LCA) in O(h)time instead of searching the entire tree.

A brute-force approach would compare subRoot against every node, which is exactly what this recursive solution does. Since each comparison uses the optimal Same Tree algorithm, this is the standard interview solution and is accepted by virtually every interviewer.

Given the roots of two binary trees p and q, determine whether they are exactly the same. Two binary trees are considered the same if: They have the same structure. Every corresponding node contains the same value

Whenever a tree problem asks you to determine something about a node based on information from its children, think post-order DFS. For this problem, every node needs: left subtree height, right subtree height, whether both subtrees are already balanced

Once you understand that the DFS is responsible only for returning subtree heights, the entire solution becomes a clean one-pass traversal of the tree. Recognise that the diameter at a node is left_height + right_height.

With recursive DFS, the answer for each node is:1 + the greater depth of its two subtrees With BFS, the answer is the number of levels processed. With iterative DFS, each node is stored together with its current depth.

Binary Tree problems are a staple in coding interviews, and Invert Binary Tree is one of the most famous ones. Despite its simplicity, this problem is an excellent exercise for understanding tree traversal and the differences between recursive DFS, iterative DFS, and BFS.