# Vertical Order Traversal

#### 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.
```

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

##### 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:
```

```
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
```

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:

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

The intuition is to find the breadth of the tree first so that we can beforehand know the maximum horizontal distance and minimum horizontal distance of a node from the root node. We can use the absolute value of minimum horizontal distance as an offset. Now we can use an array/list visited to store the visited nodes where **ith** element will store the node at** (i - **offset**”)** distance horizontally from the root node. This way we can reduce the time complexity of inserting and accessing the nodes.

Let us define a function

getBreadth(TreeNode<int>* root, int hrDistance, int &minLeft,

int &maxRight)

Which finds the minimum horizontal distance and maximum horizontal distance and stores it in **minLeft** and **rightLeft** variables respectively. And **hrDistance** stores the horizontal distance between current node and the root node.

Now consider the following steps to implement this function:

- If root is NULL then return.
- Otherwise
- Recur for left subtree i.e. getBreadth(root->left, hrDistance-1, minLeft, maxRight)
- Similarly recur for right subtree i.e. getBreadth(root->right, hrDistance+1, minLeft, maxRight)
- Now update the “minLeft” and “maxRight” with the horizontal distance of the current node.
**minLeft = min(minLeft, hrDistance)****maxRight = min(maxRight, hrDistance)**

After getting the minimum horizontal distance and maximum horizontal distance, we can create an array/list **visited** of (maxRight - minLeft + 1) size to store the nodes. Also we can set the “offset” to negative of minimum horizontal distance.

We will follow the same approach as mentioned in approach 2 to visit the nodes.

Steps are as follows:

- Make a queue for traversing level by level like :
- queue< pair< TreeNode<int>*, int> > level, where the first element of the “level” is the node and the second element of the “level” is the horizontal distance of that node from the root node.

- Push the root node into the queue say “level” and make the horizontal distance 0 with the root node.
- Iterate until “level” does not become empty:
- Each time get the current node from the front of the “level” let’s say
**currNode**. And get the horizontal distance of that node, let’s say**hrDistance**. Check if**hrDistance**is already visited or not? If it is not visited yet then make it visited with the value of “currNode” else ignore it because the previous node must have visited before the current node and its level will be less then current, so the previously-stored node will hide the current node. - If the left node exists of
**currNode**, then append the left node to the queue with one less horizontal distance than the “curNode”.**level.push({currNode->left, hrDistance-1 }).**

- If the right node exists of “curr”, then append the right node to the queue with one more horizontal distance than the “curr”.
**level.push({currNode->right, hrDistance+1 }).**

- Each time get the current node from the front of the “level” let’s say
- Finally, our values of
**visited**have all the nodes which can be viewed from the top of the tree and store the whole vertical order. Store them and return the final answer.