Binary Tree Data Structure Questions

A stack can be used to keep track of the current path being explored. We can add nodes to the stack as we traverse down the tree and backtrack by popping nodes when we reach a leaf or the sum exceeds the target.

Level order serialization stores nodes level by level, making it easy to reconstruct the tree in the same level order when deserializing.

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.

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.

If the search value is not present in the BST, the search algorithm typically returns a null pointer or a designated value (like -1) to signal that the value was not found.

Full binary trees always have an odd number of nodes due to the relationship between internal nodes and total nodes (2k + 1).

Leaf nodes have no children, so their left and right child pointers will be NULL. Checking these pointers is a reliable way to identify them.

The total number of nodes in a perfect binary tree of height 'h' is calculated as 2 raised to the power of (h + 1), minus 1.

To maintain the BST property, the replacement node should be the inorder successor of the deleted node. This is efficiently found as the leftmost child of the right subtree (or the rightmost child of the left subtree, which is equivalent).

A queue's FIFO (First-In, First-Out) property naturally lends itself to Level Order Traversal, as we want to process nodes in the order they are encountered level by level.

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.

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 the worst case, we might have a skewed tree where the number of root-to-leaf paths is proportional to the number of nodes. Additionally, each path can have up to O(n) nodes, resulting in quadratic time complexity.

The key difference lies in the order in which the root node is processed relative to its left and right subtrees.

While both approaches have the same time complexity, recursive calls consume memory on the call stack, which can lead to stack overflow issues for deep trees. Iterative approaches using a stack avoid this problem.

The diameter represents the maximum distance (in terms of edges) that you can travel within the tree by starting at one node and ending at another.

In a complete binary tree, the depth of a node is related to its index in an array-based representation. The formula Depth = log2(Index + 1) accurately captures this relationship.

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

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.

The height of a perfect binary tree can be calculated as the floor of log base 2 of (n + 1), minus 1.