Validate BST

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Problem statement

You have been given a binary tree of integers with N number of nodes. Your task is to check if that input tree is a BST (Binary Search Tree) or not.

A binary search tree (BST) is a binary tree data structure which has the following properties.

• The left subtree of a node contains only nodes with data less than the node’s data.
• The right subtree of a node contains only nodes with data greater than the node’s data.
• Both the left and right subtrees must also be binary search trees.
Example :

BST1

Answer :

Level 1: 

All the nodes in the left subtree of 4 (2, 1, 3) are smaller 
than 4, all the nodes in the right subtree of the 4 (5) are 
larger than 4.

Level 2 :

For node 2:
All the nodes in the left subtree of 2 (1) are smaller than 
2, all the nodes in the right subtree of the 2 (3) are larger than 2.
For node 5:
The left and right subtrees for node 5 are empty.

Level 3:

For node 1:
The left and right subtrees for node 1 are empty.
For node 3:
The left and right subtrees for node 3 are empty.

Because all the nodes follow the property of a binary search tree, the above tree is a binary search tree.
Detailed explanation ( Input/output format, Notes, Images )
Input Format : 
The first line contains an integer 'T', which denotes the number of test cases or queries to be run. Then the test cases follow.

The first line of input contains the elements of the tree in the level order form separated by a single space.

If any node does not have a left or right child, take -1 in its place. Refer to the example below.

Example :

Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :

Image1

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null (-1).
Note : 
The above format was just to provide clarity on how the input is formed for a given tree. 

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Output Format : 
For each test case, print ‘true’ if the binary tree is a BST, else print ‘false’.

Output for every test case will be denoted in a separate line.

Note :

You are not required to print anything explicitly. It has already been taken care of. Just implement the function.
Constraints : 
1 <= T <= 100
1 <= N <= 5000
-10^6 <= data <= 10^6 and data != -1

where N is the number of nodes in the tree, T represents the number of test cases, and ‘data’ denotes data contained in the node of the binary tree.

Time Limit: 1 sec
Sample Input 1 :
2
3 1 5 -1 2 -1 -1 -1 -1
3 2 5 1 4 -1 -1 -1 -1 -1 -1
Sample Output 1 :
 true
 false
Explanation of the Sample Input1 :
Here we have 2 test cases, hence there are 2 binary trees

Test Case 1:

Test1

 Level 1: 
For node 3 all the nodes in the left subtree (1,2) are less than 3 and all the nodes in the right subtree (5) are greater than 3. 

Level 2:
For node 1:
The left subtree is empty and all the nodes in the right subtree (2) are greater than 1.  
For node 5:
Both right and left subtrees are empty.

Level 3:
For node 2, both the right and left subtrees are empty. Because all the nodes follow the property of a binary search tree, the function should return true.

Test Case 2:

Test2

For the root node, all the nodes in the right subtree (5) are greater than 3. But node with data 4 in the left subtree of node 3 is greater than 3, this does not satisfy the condition for the binary search tree. Hence, the function should return false.
Sample Input 2 :
1
1 1 2 -1 -1 -1 -1
Sample Output 2 :
false
Explanation of the Sample Input1 :
As we can see the left child of root is equal to root which violates the property of BST as the left child should be lesser than root.
Hint

Try to narrow down the minimum and maximum values for each node.

Approaches (2)
BST property

The approach is based on the fact that the value of each node in a BST is greater than the value of all the nodes in the left subtree and smaller than the value of all the nodes in the right subtree.

 

Here is the complete algorithm-

  • For each node, we store the minimum and maximum value allowed for that node. Initially, for the root node, the minimum value would be -10^9 and the maximum value should be 10^9 (as all the integer values are allowed).
  • If the value of that node is not in the bounded range of minimum and maximum value, then return false.
  • For the left subtree of a node with data ‘x’, update the maximum value to ‘x’.
  • For the right subtree of a node with data ‘x’, update the minimum value to ‘x’.
    • This will make sure every node in the left subtree will be less than ‘x’ and every node in the right subtree will be greater than ‘x’.
  • Check this for each node of the tree, if every node is in the range then the tree is a binary search tree.
Time Complexity

O(N), where N is the number of nodes in the binary tree.

 

We are recursively traversing through all the nodes of the tree.

Space Complexity

O(N), where N is the number of nodes in the binary tree.

 

O(H), recursion stack space is used by the algorithm where H is the height of BST. In the worst case (for skewed trees), H will become N. Thus, the overall space complexity is O(N).

Code Solution
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Validate BST
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