Check if each internal node of a BST has exactly one child

Given Preorder traversal of a BST, check if each non-leaf node has only one child. Assume that the BST contains unique entries.

Examples :

Input: pre[] = {20, 10, 11, 13, 12}
Output: Yes
The given array represents following BST. In the following BST, every internal
node has exactly 1 child. Therefore, the output is true.
 20
 /
 10
 \
 11
 \
 13
 /
 12

In Preorder traversal, descendants (or Preorder successors) of every node appear after the node. In the above example, 20 is the first node in preorder and all descendants of 20 appear after it. All descendants of 20 are smaller than it. For 10, all descendants are greater than it. In general, we can say, if all internal nodes have only one child in a BST, then all the descendants of every node are either smaller or larger than the node. The reason is simple, since the tree is BST and every node has only one child, all descendants of a node will either be on left side or right side, means all descendants will either be smaller or greater.

Approach 1 (Naive):
This approach simply follows the above idea that all values on right side are either smaller or larger. Use two loops, the outer loop picks an element one by one, starting from the leftmost element. The inner loop checks if all elements on the right side of the picked element are either smaller or greater. The time complexity of this method will be O(n^2).

Approach 2:
Since all the descendants of a node must either be larger or smaller than the node. We can do the following for every node in a loop. 

1. Find the next preorder successor (or descendant) of the node. 
2. Find the last preorder successor (last element in pre[]) of the node. 
3. If both successors are less than the current node, or both successors are greater than the current node, then continue. Else, return false.

Below is the implementation of the above approach:

Yes

Time Complexity: O(n), where n is the length of the given pre[] array.
Auxiliary Space: O(1)

Approach 3 :

1. Scan the last two nodes of preorder & mark them as min & max. 
2. Scan every node down the preorder array. Each node must be either smaller than the min node or larger than the max node. Update min & max accordingly. 

Below is the implementation of the above approach:

Yes

Time Complexity: O(n), where n is the length of the given pre[] array.
Auxiliary Space: O(1)

Approach 4:

1. Create a function hasOnlyOneChild that takes an integer array pre and its size size as input.
2. Initialize min and max as the last two elements of the pre array. This is because the last element in the preorder traversal must be either the rightmost element in the tree or its ancestor.
3. For each element in the pre array starting from size - 3 to 0, do the following:                                                                                                                   a) If the element is less than the min value, update min to the element.
   b) If the element is greater than the max value, update max to the element.
   c) If the element is between min and max, return false, since the node has more than one child.
   d) If the loop completes without returning false, then all nodes have only one child, so return true.
4. In the main function, initialize an integer array pre and its size size.
5. Call the hasOnlyOneChild function with pre and size as arguments and store the returned value in a boolean variable.
6. If the boolean value is true, print "Yes". Otherwise, print "No".
7. End the program.

Yes

Time complexity: O(n)

Auxiliary Space: O(n)