## Disordered Binary Search Trees

### March 10, 2017

As in the past, a binary tree is a list of three nodes with the value in the `car`

, the left child in the `cadr`

, and the right child in the `caddr`

.

A null tree is a binary search tree. A non-null tree is a binary search tree if its left child is a binary search tree, its right child is a binary search tree, its node value is greater the head of its left child, and its node value is less than the head of its right child:

(define (ordered? lt? tree) (or (null? tree) (and (ordered? lt? (cadr tree)) (ordered? lt? (caddr tree)) (or (null? (cadr tree)) (lt? (caadr tree) (car tree))) (or (null? (caddr tree)) (lt? (car tree) (caaddr tree))))))

Here are two examples:

> (ordered? < '(4 (0 () (1 () (2 () (3 () ())))) (5 () (6 () ())))) #t > (ordered? < '(6 (0 () (1 () (2 () (3 () ())))) (5 () (4 () ())))) #f

You can run the program at http://ideone.com/i7MGDU.

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@programmingpraxis: If I understand your definition of a BST given in the solution, then the following tree would be a BST:

├── 10 (root)

│ ├── 40

│ └── 5

│ ├── 20

│ └── 1

The tree at the left (with root value 5 and children 1 and 20) is a BST, but the total tree is not, as in the left tree the value 20 is higher than 10.

Below an iterative solution in Python. Recursive is not a good idea for this in Python, because it would not work for large trees.

Unfortunately the tree is not printing well (it does in “preview”). The tree is:

root: 10 with children 5 and 40

left child of root: 5 with children 1 and 20

ANother try to print the tree

An in-order traversal of a valid binary search tree will visit the nodes in sorted order. An in-order traversal of an invalid binary search tree will visit the nodes in unsorted order.

Here’s a solution in Java that validates a binary search tree by considering whether nodes visited in-order are sorted. Morris traversal is used for O(1) space, O(n) runtime.

The example checks a valid and invalid tree.

Output: