Maximum Difference Between Node And Ancestor

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Average time to solve is 10m
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IBMLeena AI

Problem statement

Given a binary tree, return the maximum absolute difference between a node and its ancestor, for any ancestor-node pair in the binary tree.

The ancestor of any node is defined as the node/nodes which come above the current node in the binary tree. For example, the root node is ancestor of every node in the binary tree.

For example :

TC1(2)

In the above figure, we have many nodes which have a node and ancestor relationship.

Some of them are, and their difference is:

|1-4|=3
|2-4|=2
|3-4|=1
|6-4|=2
|7-4|=3
And more

Out of all the possible parent ancestor pairs, the one with the maximum absolute difference is between nodes (7 and 4) and (1 and 4) which is 3. Therefore, the answer to this case is 3.

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 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 on its place.

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

example

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, return a single integer denoting the maximum difference between node and ancestor
Note : 
Do not print anything, just return the maximum difference between the node and the ancestor
Constraints : 
1 <= T <= 50
1<= N <= 10^3

Where 'N' denotes the number of nodes in the binary tree.

Time limit: 1 second
Sample Input 1 :
2
6 4 7 -1 -1 -1 8 9 -1 -1 -1 
4 1 5 2 -1 6 -1 -1 3 -1 7 -1 -1 -1 -1
Sample Output 1 :
3
3
Explanation Of Sample Input 1 :
Test Case 1:
The tree is:

maxDiff

We can clearly see that the maximum difference pair between node and ancestor is the 6 and 9 pair whose difference is (|6-9|=3). Therefore, we return 3. Note that any other pair lets take for example 8 and 9 gives a difference of (|8-9| = 1 ), which is not the maximum.


Test Case 2:
The Tree is:

top wala

The maximum difference pair, in this case, are the nodes 4 and 1 and their difference is (|4-1|=3) and nodes 4 and 7 whose difference is (|7-4|=3) , both of which give a difference of 3. Therefore we return 3.
Sample Input 2 :
2
1 2 3 -1 -1 5 6 7 8 -1 -1 -1 -1 -1 -1
20 8 22 5 3 -1 25 -1 -1 10 14 -1 -1 -1 -1 -1 -1
Sample Output 2 :
7
17
Hint

Recursively try every pair of nodes.

Approaches (2)
Brute Force Approach
  • The idea is to traverse the binary tree and for each node check all its children and find the difference of the node with each of its children.
  • This can be done by recursively checking for each node the maximum and minimum node in its left and right subtree and return its maximum difference over all nodes in the tree.

Keeping in mind the above idea, we can devise the following recursive  approach:

  • Define 2 functions ‘maxInSub’ and ‘minInSub’ to calculate the maximum and minimum in any subtree for every node.
  • We can calculate both the functions recursively by comparing the max or min with the current node and recursively calculating answer for the left and right subtree.
  • Then, recursively check for each node for the maximum in its subtree and the minimum in its subtree and find the maximum difference.
  • Finally, return the overall maximum difference value over all nodes of the tree.
Time Complexity

O(N^2), where ‘N’ is the number of nodes in the binary tree.

 

We are recursively traversing through all the nodes of the binary tree, and for each node we calculate the maximum and minimum in its left and right subtree which takes time of the order of ‘N’. Hence, the overall complexity is the order of N^2.

Space Complexity

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

 

Stack takes space for recursive calls of the order ‘N’.

Code Solution
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Maximum Difference Between Node And Ancestor
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