Predecessor and Successor In BST

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Problem statement

Given a binary search tree of integers containing 'N' nodes. You have also been given an integer X.

Your task is to find the inorder successor and predecessor of the given X. Formally, print an array/list containing the inorder predecessor and successor in the same order.

For Example: 
For the BST given below:

alttext

The inorder predecessor of 6 is 4.
The inorder successor of 6 is 7.
The inorder predecessor of 10 is 8.
The inorder successor of 10 is 13.

Note:

If there is no inorder predecessor or successor of 'X', then add -1 to the answer vector in its place.
Detailed explanation ( Input/output format, Notes, Images )
Input Format: 
The first line of input contains a single integer 'T', representing the number of test cases or queries to be run. 

Then the 'T' test cases follow:

The first line of each test case contains elements in the level order form. 
The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place. Refer example given below for more clarity.

The second line of each test case contains a single integer 'X'.

Example:

The input for the tree depicted in the below image will be:

alt text

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: 
Print a single line containing the inorder predecessor and successor of 'X' separated by a single space.

The output of each test case is printed in a separate line.

Note:

You do not need to print anything, it has already been taken care of. Just implement the given function.
Constraints: 
1 <= T <= 5
1 <= N <= 5000
0 <= DATA <= 10^9
0 <= X <= 10^9

Where 'DATA' is the value of any node in the BST.

Time Limit: 1sec
Sample Input 1:
1
8 5 10 2 6 -1 -1 -1 -1 -1 7 -1 -1
5
Sample Output 1:
2 6

Explanation of the Sample Input 1:

For the given ‘X’ = 5, according to the inorder view, 2 is the parent of the left subtree and 6 is the parent of the right subtree, hence 2 and 6 are inorder predecessors and successor respectively.
Sample Input 2:
2
8 5 10 2 6 -1 -1 -1 -1 -1 7 -1 -1
6
8 5 10 2 6 -1 -1 -1 -1 -1 7 -1 -1
2
Sample Output 2:
5 7
-1 5
Approaches (1)
Using In-order Traversal

The main idea is to do an inorder traversal and try to find the key. If we find the key then the maximum value in the left subtree is the predecessor and the minimum value in the right subtree is the successor of the given key ‘X’.

 

  • Maintain two variables ‘PRE’ and ‘SUC’ which point to the predecessor and successor respectively.
  • We use a helper function ‘FIND_PRE_SUC_HELP’, which take the following as parameter :
    • ‘ROOT’
    • ‘PRE’
    • ‘SUC’
    • ‘KEY’ i.e. ‘X’
  • If the current node's data is equal to ‘KEY’ then :
    • The maximum value in the left subtree is the predecessor.
    • The minimum value in the right subtree is the successor.
  • If after exiting ‘FIND_PRE_SUC_HELP’ any of ‘PRE’ or ‘SUC’ are NULL then that value was not found then add -1 to the vector.
  • Else add the value pointed by ‘PRE’ and then ‘SUC’.
Time Complexity

O(N), where ‘N’ is the number of nodes in the given tree.

 

Since we are traversing the tree once, therefore the net time complexity will be O(N).

Space Complexity

O(N), where ‘N’ is the number of nodes in the given tree.

 

Since we are traversing the tree using recursion and a call stack of the tree’s height would be made and in the worst case that would be equal to ‘N’, therefore O(N).

Code Solution
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Predecessor and Successor In BST
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