Diagonal Traversal

Moderate
0/80
Average time to solve is 15m
28 upvotes
Asked in company
Qualcomm

Problem statement

You are given a binary tree having ‘N’ nodes. Your task is to return its diagonal traversal.

A binary tree is a hierarchical data structure in which each node has at most two children.

For Example: 

For the above binary tree, the output will be:
8 10 14
3 6 7 13
1 4
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 each test case 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 :


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 each diagonal traversal in a new line.
The output of each test case should be printed in a separate line.
Note 
You are not required to print anything, it has already been taken care of. Just implement the function.
Constraints: 
1 <= T <= 100
1 <= N <= 3 * 10^3
1 <= nodeVal <= 10^9

Time Limit: 1 sec.
Sample Input 1:
2
3 1 2 4 -1 -1 -1 -1 -1
1 2 3 4 -1 5 -1 -1 -1 -1 6 -1 -1
Sample Output 1:
3 2 
1 
4
1 3 
2 5 6 
4
Explanation For Sample Output 1:
For the first test case, the tree will be:


So, the output will be:
3 2
1
4

For the second test case, the tree will be:


So, the output will be:
1 3
2 5 6
4
Sample Input 2:
2
1 2 -1 -1 -1
1 3 2 -1 -1 -1 -1
Sample Output 2:
1 
2 
1 2 
3
Hint

Try to find the relationship between the diagonal level of the current node with the left and right child.

Approaches (1)
Recursion

We will implement a recursive ‘helper’ function to traverse the tree. Note that the right child of a node will have the same diagonal level as the parent node and the left child will have the next diagonal level as that of the parent node.

The ‘helper’ function will take a node, diagonal level of the node, and a map as arguments and will store the current node in the map corresponding to its diagonal level.
 

In the ‘diagonal’ function we will call the ‘helper’ function and then store the information in a vector of vectors from the map, and finally return it.

 

Algorithm:

 

  1. In the ‘helper’ function:
    1. We will pass the root, diagonal level, and a map to this function.
    2. It will store the current node to its corresponding diagonal level and call the ‘helper’ function for the left and right child.
    3. The left child will have the next diagonal level and the right child will have the same diagonal level as that of the parent node.

 

  1. In the given function:
    1. we will just declare a map of int and vector to store the elements in their corresponding diagonal level.
    2. Then we will call the ‘helper’ function.
    3. Finally, we will store the information to a vector from the map and will return it.
Time Complexity

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

Since we will visit each node exactly once and we are using a map to store the information, the total time complexity will be O(N*log(N)).

Space Complexity

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

 

Since we are using an extra map to store the information.

Code Solution
(100% EXP penalty)
Diagonal Traversal
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