Print a Binary Tree in Vertical Order
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

Question

Print a Binary Tree in Vertical Order

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}.

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

Accepted Answer

Level Order Traversal can be used here.
Algorithm :
1. Maintain a hash map for the branch of each node.
2. Traverse the tree in level order fashion.
3. In level order traversal, maintain a queue which holds, node and its vertical branch.
* pop from queue.
* add this node's data in vector corresponding to its branch in the hash.
* if this node hash left child, insert in the queue, left with branch - 1.
* if this node hash right child, insert in the queue, right with branch + 1.

Time Complexity of the above implementation is O(n Log n).

Accepted Answer

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:

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.

Space Complexity: O(n)Explanation:

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).

Time Complexity: O(nlogn)Explanation:

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).

C++ (g++ 5.4)

/*
    Time Complexity: O(N log N)
    Space Complexity: O(N)

    Where 'N' is the number of nodes in a binary tree.
*/

#include 

vector verticalOrderTraversal(TreeNode *root)
{
    // Store the final result.
    vector result;

    // Base case
    if (root == NULL)
    {
        return result;
    }

    // Create a map and store vertical order in map.
    map> store;

    // Initial horizontal distance.
    int hd = 0;

    // Every pair of queue contains node and horizontal distance with respect to the root node.
    queue *, int>> level;

    level.push(make_pair(root, hd));

    while (level.empty() == false)
    {
        // Pop element from the queue.
        pair *, int> temp = level.front();
        level.pop();

        hd = temp.second;

        TreeNode *node = temp.first;

        // Insert the value of current node in the vector of map.
        store[hd].push_back(node->data);

        // For the left subtree, we pass the HD as the Horizontal distance of root minus 1.
        if (node->left != NULL)
        {
            level.push(make_pair(node->left, hd - 1));
        }

        // For the right subtree, we pass the HD as Horizontal Distance of root plus 1.
        if (node->right != NULL)
        {
            level.push(make_pair(node->right, hd + 1));
        }
    }

    // Traverse the map and store node values at every horizontal distance (hd).
    map>::iterator it;

    for (it = store.begin(); it != store.end(); it++)
    {
        for (int i = 0; i < it->second.size(); ++i)
        {
            result.push_back(it->second[i]);
        }
    }

    // Return the vertical order traversal of the given binary tree.
    return result;
}

Java (SE 1.8)

/*
    Time Complexity - O(N)
    Space Complexity - O(N)

    Where N is the number of nodes in the Binary Tree.
*/

import java.util.Map;
import java.util.Queue;
import java.util.TreeMap;
import java.util.ArrayList;
import java.util.LinkedList;

class Pair
{
    public TreeNode root;
    public int hDistance;

    // Constructor
    Pair(TreeNode root, int hDistance)
    {
        this.root = root;
        this.hDistance = hDistance;
    }
}

public class Solution
{
    public static ArrayList verticalOrderTraversal(TreeNode root)
    {
        // Store the final result.
        ArrayList result = new ArrayList();

        // Base case
        if (root == null)
        {
            return result;
        }

        // Create a map and store vertical order in map.
        Map> store = new TreeMap>();

        // Initial horizontal distance.
        int hd = 0;

        // Every pair of queue contains node and horizontal distance with respect to the root node.
        Queue level = new LinkedList();

        level.add(new Pair(root, hd));

        while (level.isEmpty() == false)
        {
            // Pop element from the queue.
            Pair temp = level.remove();

            hd = temp.hDistance;

            TreeNode node = temp.root;

            // Insert the value of current node correspond to hd.
            ArrayList element = store.get(hd);
            if (element == null)
            {
                element = new ArrayList();
            }
            element.add(node.data);
            store.put(hd, element);

            // For the left subtree, we pass the HD as the Horizontal distance of root minus 1.
            if (node.left != null)
            {
                level.add(new Pair(node.left, hd - 1));
            }

            // For the right subtree, we pass the HD as Horizontal Distance of root plus 1.
            if (node.right != null)
            {
                level.add(new Pair(node.right, hd + 1));
            }
        }

        // Traverse the map and store node values at every horizontal distance (hd).
        for (Map.Entry> it : store.entrySet())
        {
            for (int i = 0; i < it.getValue().size(); ++i)
            {
                result.add(it.getValue().get(i));
            }
        }

        // Return the vertical order traversal of the given binary tree.
        return result;
    }
}

Python (3.5)

'''
    Time Complexity: O(N log N)
    Space Complexity: O(N)

    Where 'N' is the number of nodes in a binary tree.
'''

from collections import deque


def verticalOrderTraversal(root):

    # To store the final result
    result = []

    # Base Case
    if root == None:
        return result

    # Create a dictionary and store vertical order in map.
    store = {}
    # Initial horizontal distance.
    hd = 0
    # Every pair of deque contains node and horizontal distance with respect to the root node.
    level = deque()
    level.append((root, hd))

    while len(level):
        # Pop element from the queue.
        temp = level[0]
        level.popleft()

        hd = temp[1]
        node = temp[0]

        # Insert the value of current node in the vector of map.
        if hd not in store:
            store[hd] = []
        store[hd].append(node.data)

        # For the left subtree, we pass the HD as the Horizontal distance of root minus 1.
        if node.left != None:
            level.append((node.left, hd - 1))
        # For the right subtree, we pass the HD as Horizontal Distance of root plus 1.
        if node.right != None:
            level.append((node.right, hd + 1))

    # Traverse the map and store node values at every horizontal distance (hd).
    for it in sorted(store.keys()):
        for i in store[it]:
            result.append(i)

    # Return the vertical order traversal of the given binary tree.
    return result

Accepted Answer

Efficient Approach

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* 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
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:

Make a queue for traversing level by level like :
1. queue< pair< TreeNode*, 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:
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 }).
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.

Space Complexity: O(n)Explanation:

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).

Time Complexity: O(n)Explanation:

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).

C++ (g++ 5.4)

/*
    Time Complexity : O(N)
    Space Complexity : O(N)
    Where 'N' is the number of nodes in the given binary tree.
*/

#include 

void getBreadth(TreeNode *root, int hrDistance, int &minLeft, int &maxRight)
{
    if (root == NULL)
    {
        return;
    }

    // Recur for the left subtree.
    getBreadth(root->left, hrDistance - 1, minLeft, maxRight);

    // Recur for the right subtree.
    getBreadth(root->right, hrDistance + 1, minLeft, maxRight);

    // Update the 'minLeft' and 'maxRight' with the horizontal distance of the current node.
    minLeft = min(minLeft, hrDistance);
    maxRight = max(maxRight, hrDistance);
}

vector verticalOrderTraversal(TreeNode *root)
{
    // Store the final result.
    vector result;

    // Base case
    if (root == NULL)
    {
        return result;
    }

    //  To store the minimum and maximum horizontal distance.
    int minLeft = 0, maxRight = 0;

    //  Get the minimum and maximum horizontal distance.
    getBreadth(root, 0, minLeft, maxRight);

    // Create a map and store vertical order in map.
    vector> store(maxRight - minLeft + 1, vector());

    // Initial horizontal distance.
    int hd = 0;
    int offset = abs(minLeft);

    // Every pair of queue contains node and horizontal distance with respect to the root node.
    queue *, int>> level;

    level.push(make_pair(root, hd));

    while (level.empty() == false)
    {
        // Pop element from the queue.
        pair *, int> temp = level.front();
        level.pop();

        hd = temp.second;

        TreeNode *node = temp.first;

        // Insert the value of current node in the vector of map.
        store[hd + offset].push_back(node->data);

        // For the left subtree, we pass the HD as the Horizontal distance of root minus 1.
        if (node->left != NULL)
        {
            level.push(make_pair(node->left, hd - 1));
        }

        // For the right subtree, we pass the HD as Horizontal Distance of root plus 1.
        if (node->right != NULL)
        {
            level.push(make_pair(node->right, hd + 1));
        }
    }

    for (int i = 0; i < store.size(); i++)
    {
        for (int j = 0; j < store[i].size(); ++j)
        {
            result.push_back(store[i][j]);
        }
    }

    // Return the vertical order traversal of the given binary tree.
    return result;
}

Python (3.5)

'''
    Time Complexity: O(N)
    Space Complexity: O(N)

    Where 'N' is the number of nodes in a binary tree.
'''

from collections import deque


def getBreadth(root, hrDistance, minLeft, maxRight):
    if root == None:
        return minLeft, maxRight

    # Recur for the left subtree.
    minLeft, maxRight = getBreadth(root.left, hrDistance - 1, minLeft, maxRight)

    # Recur for the right subtree.
    minLeft, maxRight = getBreadth(root.right, hrDistance + 1, minLeft, maxRight)

    # Update the 'minLeft' and 'maxRight' with the horizontal distance of the current node.
    minLeft = min(minLeft, hrDistance)
    maxRight = max(maxRight, hrDistance)

    return minLeft, maxRight


def verticalOrderTraversal(root):
    # To store the final result
    result = []

    # Base Case
    if root == None:
        return result

    # To store the minimum and maximum horizontal distance.
    minLeft, maxRight = 0, 0
    # Get the minimum and maximum horizontal distance.
    minLeft, maxRight = getBreadth(root, 0, minLeft, maxRight)

    # Create a map from a 2D list and store vertical order in map.
    store = [[] for i in range(maxRight - minLeft + 1)]

    # Initial horizontal distance.
    hd = 0
    offset = abs(minLeft)
    # Every pair of queue contains node and horizontal distance with respect to the root node.
    level = deque()
    level.append((root, hd))

    while len(level):
        # Pop element from the queue.
        temp = level[0]
        level.popleft()
        hd = temp[1]
        node = temp[0]
        # Insert the value of current node in the vector of map.
        store[hd + offset].append(node.data)

        # For the left subtree, we pass the HD as the Horizontal distance of root minus 1.
        if node.left != None:
            level.append((node.left, hd - 1))
        # For the right subtree, we pass the HD as Horizontal Distance of root plus 1.
        if node.right != None:
            level.append((node.right, hd + 1))

    # Traverse the map and store node values at every horizontal distance (hd).
    for it in range(0, len(store)):
        for i in store[it]:
            result.append(i)

    # Return the vertical order traversal of the given binary tree.
    return result

Java (SE 1.8)

/*
    Time Complexity : O(N)
    Space Complexity : (N)
    Where 'N' is the number of nodes in the given binary tree.
*/

import java.util.Queue;
import java.util.ArrayList;
import java.util.LinkedList;

class Pair
{
    TreeNode first;
    int second;

    Pair(TreeNode first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

public class Solution
{

    private static void getBreadth(TreeNode root, int hrDistance, int []minLeft, int []maxRight)
    {
        if (root == null)
        {
            return;
        }

        //  Recur for the left subtree.
        getBreadth(root.left, hrDistance - 1, minLeft, maxRight);

        //  Recur for the right subtree.
        getBreadth(root.right, hrDistance + 1, minLeft, maxRight);

        //  Update the "minLeft" and "maxRight" with the horizontal distance of the current node.
        minLeft[0] = Math.min(minLeft[0], hrDistance);
        maxRight[0] = Math.max(maxRight[0], hrDistance);
    }

    public static ArrayList verticalOrderTraversal(TreeNode root)
    {

        //  For storing top view element
        ArrayList result = new ArrayList();

        if (root == null)
        {
            return result;
        }

        //  To store the minimum and maximum horizontal distance.
        int []minLeft = {0};
        int []maxRight = {0};
        int hd = 0;

        //  Get the minimum and maximum horizontal distance.
        getBreadth(root, 0, minLeft, maxRight);

        //  To store the nodes.
        ArrayList> store = new ArrayList>(maxRight[0] - minLeft[0] + 1);

        for (int i = 0 ; i < (maxRight[0] - minLeft[0] + 1); i++)
        {
            store.add(new ArrayList());
        }

        //  Offset value.
        int offset = Math.abs(minLeft[0]);

        //  For iterating level by level
        Queue level = new LinkedList();

        //  Append root to the level with 0 horizontal distance
        level.add(new Pair(root, 0));

        //  Iterate untill level does not become empty

        while (level.isEmpty() == false)
        {
            // Pop element from the queue.
            Pair temp = level.remove();

            hd = temp.second;

            TreeNode node = temp.first;

            // Insert the value of current node in the vector of map.
            store.get(hd + offset).add(node.data);

            // For the left subtree, we pass the HD as the Horizontal distance of root minus 1.
            if (node.left != null)
            {
                level.add(new Pair(node.left, hd - 1));
            }

            // For the right subtree, we pass the HD as Horizontal Distance of root plus 1.
            if (node.right != null)
            {
                level.add(new Pair(node.right, hd + 1));
            }
        }

        for (int i = 0; i < store.size(); i++)
        {
            for (int j = 0; j < store.get(i).size(); ++j)
            {
                result.add(store.get(i).get(j));
            }
        }

        // Return the vertical order traversal of the given binary tree.
        return result;
    }
}

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:

For example:

Input Format:

Explanation :

Note :

Output Format:

Note:

Constraints:

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