Vertical Order Traversal

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:

1. If root is NULL then return.
2. Otherwise
   1. Recur for left subtree i.e.  getBreadth(root->left, hrDistance-1, minLeft, maxRight)
   2. Similarly recur for right subtree i.e. getBreadth(root->right, hrDistance+1, minLeft, maxRight)
   3. Now update the “minLeft” and “maxRight” with the horizontal distance of the current node.
      1. minLeft = min(minLeft, hrDistance)
      2. 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:

1. Make a queue for traversing level by level like :
   1. 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.
2. Push the root node into the queue say “level” and make the horizontal distance 0 with the root node.
3. Iterate until “level” does not become empty:
   1. 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.
   2. If the left node exists of currNode, then append the left node to the queue with one less horizontal distance than the “curNode”.
      1. level.push({currNode->left,  hrDistance-1 }).
   3. If the right node exists of “curr”, then append the right node to the queue with one more horizontal distance than the “curr”.
      1. level.push({currNode->right, hrDistance+1 }).
4. 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.

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

Since we are using a post-order traversal to get the breadth of the tree and then traversing at every node and storing the horizontal distance of every node into an array/list, So inserting an element into the array/list will take O(1) time, and in the worst case (Skewed Trees) there will be ‘N’ distinct horizontal distance so it will take O(N).

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

We are storing the top view of the tree. So in the worst case (Skewed Trees), all nodes of the given tree will be the top view elements, and also there will be N distinct horizontal distance to be stored. So overall space complexity will be O(N).