Last Updated: 11 Oct, 2020
Diagonal Anagram
Moderate
Asked in companies
You are given two Binary Trees. You need to return true if the diagonals of trees are anagram to each other, otherwise you need to return false.
Diagonals of the binary tree are :
There are three diagonals :
Diagonal 1 : 8 10 14
Diagonal 2 : 3 6 7 13
Diagonal 3 : 1 4
For example:
For the given binary trees:
Output: True
Explanation: There are 3 diagonals in each tree.
Tree1:
Diagonal 1: 5 10 9
Diagonal 2: 6 3 5
Diagonal 3: 2
Tree2:
Diagonal 1: 5 9 10
Diagonal 2: 3 6 5
Diagonal 3: 2
Since diagonal 1 of tree 1 is an anagram of diagonal 1 of tree 2, similarly diagonal 2 of tree 1 is an anagram of diagonal 2 of tree 2 and similarly with diagonal 3.
Thus, its output should be True.
Input Format:
The first line of the test case contains elements of the first binary tree 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 on its place.
The second line of the test case contains elements of the second binary tree 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 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).
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:
The only line of output will contain “True” or “False”.
Note:
You are not required to print the output explicitly, it has already been taken care of. Just implement the function.
Constraints:
0 <= N, M <= 10^5
0 <= node data <= 10^9
Time Limit: 1sec
Approaches
To add every diagonal on the map.
- To store all diagonals separately, the hash map is used with its key as a diagonal number and value as the list containing the elements in that diagonal.
- Assign a distance 0 to the root.
- Create a class MAX_SLOPE that will store the maximum slope of the tree i.e total no of diagonal.
- Run a helper function with parameter root, distance as 0, and map.
- In the helper function,
- Check if the root is null, then return.
- Update the MAX_SLOPE to a maximum of MAX_SLOPE and d.
- Update the list of that particular diagonal by adding the current node data.
- Recur for left by incrementing D by 1.
- Recur for the right but do not change D.
- Finally, check for anagrams, if all the diagonals are anagram to each other then return true otherwise return false.
For checking anagram
- If MAX_SLOPE of both the trees are not the same then return false.
- For every slope of the tree, get the list from the map and check for the anagram. If they are not anagram then return false.
- Create two maps for both diagonals.
- Store all the elements of the list in the map with its frequency.
- Check if the frequency of all the keys is the same in both the map then return true otherwise return false.
- To store diagonals for each diagonal separately, a Hash map is used with its key as a diagonal number and value as the list containing the elements in that diagonal.
- Assign a distance 0 to the root.
- To store the vertical distance, create another map NODE_DIST where the binary tree node will be the key and vertical distance i.e d would be the value.
- Create a class MAX_SLOPE that will store the maximum slope of the tree i.e total no of diagonal.
- Create a queue for level order traversal.
- Push the root node into the queue and store its vertical distance 0 into the map NODE_DIST.
- While the queue is not empty
- Get the front node of the queue as the current node.
- While the current node is not null
- Add the element of the node to its corresponding map key.
- If the left child of the node is not null then add the left child into the queue and store the vertical distance D+1 in NODE_DIST.
- Update the current node to the right of the current node.
- Finally, check for anagrams, if all the diagonals are anagram to each other then return true otherwise return false.
For checking anagram:
- If MAX_SLOPE of both the trees are not the same then return false.
- For every slope of the tree, get the list from the map and check for the anagram. If they are not anagram then return false.
- Create two maps for both diagonals.
- Store all the elements of the list in the map with its frequency.
- Check if the frequency of all the keys is the same in both the map then return true otherwise return false.