Binary Tree Data Structure Questions

Level Order Traversal explores the tree level by level, visiting all nodes at a particular level before moving to the next, which aligns with the Breadth First Search strategy.

Preorder traversal visits the root first (1), then the left subtree (2, 4, 5), and finally the right subtree (3).

The first stack helps in the general traversal, while the second stack is used to reverse the order in which nodes are processed, effectively giving us the postorder sequence.

The worst-case scenario occurs when the BST is essentially a linear linked list (all nodes are added in a strictly ascending or descending order). In this case, each insertion requires traversing the entire tree, leading to O(n) time complexity.

Perfect Binary Trees are the most rigidly structured, demanding complete filling of all levels and uniform leaf node depth.

Heap Sort leverages the balanced nature of (near) perfect binary trees for efficient sorting in O(n log n) time.

A simple preorder or postorder serialization without null markers cannot uniquely represent a binary tree. We need additional information, such as null markers or a combination of preorder and inorder traversals, to reconstruct the original tree accurately.

Complete Binary Trees have a strict structure where all levels are filled except potentially the last one, and nodes are filled from left to right on the last level.

The order in which Preorder Traversal visits nodes can be used to reconstruct the original tree. This is helpful in creating a copy where nodes are different entities but maintain the same structure.

Postorder traversal is most suitable because it processes the left subtree, right subtree, and then the node itself. This allows us to determine if both nodes are present in either subtree before processing the current node.

With an efficient approach (like using DFS to calculate heights along with diameter), you can find the diameter in linear time, O(n), where n is the number of nodes.

In a complete binary tree, all levels are completely filled except possibly the last level. This balanced structure leads to a logarithmic relationship between the number of nodes (n) and the height, where height is floor(log2(n)).

The recursive approach incurs space overhead due to the function call stack. In the worst case, the tree might be skewed, leading to a recursion depth proportional to the number of nodes.

Postorder traversal (left subtree, right subtree, root) is suitable for deleting all nodes because it ensures that a node is deleted only after its children have been deleted.

Serialization allows us to represent the tree as a linear sequence of data that can be easily stored and retrieved from persistent storage, making it useful for saving and loading tree structures.

In a full binary tree, every internal node has exactly two children. This relationship leads to the property that the number of leaf nodes is always one more than the number of internal nodes.

In a binary tree, each node can have at most two children. Therefore, the maximum number of nodes at a given level 'l' doubles with each level, leading to 2^l nodes.

To calculate the height, we potentially need to visit all nodes in the worst case (e.g., a skewed tree). Hence, the time complexity is O(n), where n is the number of nodes.

In a BST, Inorder Traversal visits nodes in ascending order due to the property of BSTs where the left subtree has smaller values than the root, and the right subtree has larger values.

The height of a node is the longest path from that node to a leaf node (counting edges). The depth of a node is the number of edges from the root to that node.