Check whether BST contains Dead End or not

Given a Binary search Tree that contains positive integer values greater than 0. The task is to check whether the BST contains a dead end or not. Here Dead End means, we are not able to insert any element after that node.

Examples: 

Input: root[] = [8, 5, 9, 2, 7, N, N, 1]
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Output: true
Explanation: Node 1 is a Dead End in the given BST.

Input: root[] = [8, 7, 10, 2, N, 9, 13]
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Output: true
Explanation: Node 9 is a Dead End in the given BST.

Observation:

A dead end in a Binary Search Tree (BST) refers to a leaf node where no further insertions can be made without violating the properties of the BST. Since BSTs typically contain only positive integers greater than 0, a node becomes a dead end when both of its adjacent values that is, (value - 1) and (value + 1) — are already present in the tree. In such cases, no new node can be inserted in that position while maintaining the BST's structural rules.

Note: For value 0, we return true as this value cannot be added to the tree.

[Naive Approach] Using Recursion - O(n * h) time and O(h) space

The idea is to iterate through each leaf node and check whether the values value - 1 and value + 1 exist in the tree. If both of these values are present, it indicates a dead end, and the function returns true. This check is performed recursively while traversing the tree.

true

[Better Approach - 1] Using 2 Hash Sets - O(n) time and O(n) space

The idea is to store all node values in one set and only leaf node values in another set. Then, for each leaf node, we check if both its immediate predecessor and successor exist in the all-node set. If they do, then that leaf is a dead end because no other number can be inserted in that place in the BST.

Step by Step Implementation:

1. Traverse the BST and insert each node value into a set for all nodes.
2. During the same traversal, insert all leaf node values into a leaf set.
3. For every value in the leaf set, check if both (value - 1) and (value + 1) exist in the all-node set.
4. If such a condition is found for any leaf node, return true. Else, return false.

true

[Better Approach - 2] Using 1 Hash Set - O(n) time and O(n) space

The idea is to store all node values in one set and then traverse the tree again to identify leaf nodes. For each leaf node, check if both (value - 1) and (value + 1) exist in the same set. If yes, it indicates a dead end.

true

[Optimized Approach] Using Recursion and Range Values - O(n) time and O(h) space

The idea is to perform depth first search traversal on the tree, while maintaining the range of each subtree root. If for any leaf node, the size of range becomes 1 (this value is taken by leaf node already), it means no node can be inserted to this leaf node, and hence it is a dead end.

Step by step approach:

1. Perform Depth first search on the root node and initialize the range as [1, Integer Maximum].
2. For each node:
   • If node is null, return false.
   • If node is leaf node and the range of node is 1, return true as no other node can be inserted.
   • Recur for left and right subtree.

Illustration:

true