Vertical Order Traversal

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Problem statement

Given a binary tree, return the vertical order traversal of the values of the nodes of the given tree.

For each node at position (X, Y), (X-1, Y-1) will be its left child position while (X+1, Y-1) will be the right child position.

Running a vertical line from X = -infinity to X = +infinity, now whenever this vertical line touches some nodes, we need to add those values of the nodes in order starting from top to bottom with the decreasing ‘Y’ coordinates.

Note: 
If two nodes have the same position, then the value of the node that is added first will be the value that is on the left side.
For example: 
For the binary tree in the image below.

alt text

The vertical order traversal will be {2, 7, 5, 2, 6, 5, 11, 4, 9}.
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 are as follows.

The only 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. So -1 would not be a part of the tree nodes.

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

alt text

For example taking a tree:

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 the vertical order traversal of the given binary tree separated by single spaces.

Print the output of each test case 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' <= 100
0 <= 'N' <= 3000
0 <= 'VAL' <= 10^5

Where 'VAL' is the value of any binary tree node.

Time Limit: 1 sec
Sample Input 1:
2
1 2 3 4 -1 -1 -1 -1 -1
1 -1 2 -1 -1
Sample Output 1:
4 2 1 3 
1 2
Explanation of Sample Input 1:
For the first test case, the vertical order traversal of the given binary tree will be {{4}, {2}, {1}, {3}}.

For the second test case, the vertical order traversal of the given binary tree will be {{1}, {2}}.
Sample Input 2:
2
2 1 -1 -1 -1
0 1 2 4 5 3 6 -1 -1 7 -1 -1 -1 -1 -1 -1 -1
Sample Output 2:
1 2
4 1 7 0 5 3 2 6
Explanation of Sample Input 2:
For the first test case, the vertical order traversal of the given binary tree will be {{1}, {2}}.

For the second test case, the vertical order traversal of the given binary tree will be {{4}, {1, 7}, {0, 5, 3}, {2}, {6}}.
Hint

Think of keeping a track of horizontal distances of each node of the given binary tree with respect to the root node.

Approaches (2)
Iterative Approach

Our very basic intuition is that while traversing the given binary tree, we need to keep a track of the horizontal distance of all the nodes of the given binary tree with respect to the root node.

 

We initially pass the horizontal distance as 0 for root. For the left subtree, we pass the ‘HD’ as the Horizontal distance of root minus 1. For the right subtree, we pass the ‘HD’ as Horizontal Distance of root plus 1. For every Horizontal distance value, we maintain a list of nodes into the Map. Whenever we see a node in traversal, we go to the Map entry and add the node into the Map using Horizontal distance as the key in the Map.

 

Steps are as follows:

 

  1. We maintain a Map for the branch of each node.
  2. Start traversing of the given binary tree in the level order traversal.
  3. In the level order traversal,
  4. Maintain a queue which holds a node of the given binary tree and its corresponding vertical branch.
  5. While iterating the queue until it is not empty we first pop from the queue.
  6. Add the current node’s value in the list/array corresponding to its branch into the Map.
  7. If the current node has the left child, insert it in the queue, pass horizontal distance as ‘HD’ - 1.
  8. If the current node has the right child, insert it in the queue, pass horizontal distance as ‘HD’ + 1.
Time Complexity

O(N*logN), where N is the number of nodes in the given binary tree.

 

While traversing, we visit each node of the given binary tree which results in order of N. And note that the above implementation uses the map which takes O(logN) complexity for its implementation. So the overall time complexity will be O(N*logN).

Space Complexity

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

 

We are storing all the nodes for returning the answer; thus, the answer will be the size of N, and also, we are using the level order traversal. So in the worst case, when the given binary tree will be the balanced binary tree, then the total number of nodes at the bottom level of the tree will be ((N/2)+1). So at the bottom level size of the queue will be the order of O(N). So overall space complexity will be O(N).

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