|1-4|=3
|2-4|=2
|3-4|=1
|6-4|=2
|7-4|=3
And more
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 :
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).
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
For each test case, return a single integer denoting the maximum difference between node and ancestor
Do not print anything, just return the maximum difference between the node and the ancestor
1 <= T <= 50
1<= N <= 10^3
Where 'N' denotes the number of nodes in the binary tree.
Time limit: 1 second
Keeping in mind the above idea, we can devise the following recursive approach:
For this tree, we have four sequences:
[8, 3, 2], [8, 3, 6, 4], [8, 3, 6, 4], [8, 10, 14, 13]
For each sequence, a left element is an ancestor of its successors. So the max element - the min element in each sequence is a local max difference between node and ancestor, and the global maximum should be max(max element - min element, for each sequence). Such a comparison could be done quite easily.
We use DFS to find these sequences. For a node in a sequence, we maintain the current (till we reach this node) max and min value of the sequence. When we reach a leaf node, the current max and min should be those of the whole sequence.
Keeping in mind the above idea, we can devise the following recursive approach:
Preorder Traversal
Inorder Traversal
Postorder Traversal
Height of Binary Tree
Locked Binary Tree