# Print a Binary Tree in Vertical Order | Set 2 (Map based Method)

Given a binary tree, print it vertically. The following example illustrates vertical order traversal.

```           1
/    \
2      3
/ \   /   \
4   5  6   7
/  \
8   9

The output of print this tree vertically will be:
4
2
1 5 6
3 8
7
9``` ## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We have discussed a O(n2) solution in the previous post. In this post, an efficient solution based on the hash map is discussed. We need to check the Horizontal Distances from the root for all nodes. If two nodes have the same Horizontal Distance (HD), then they are on the same vertical line. The idea of HD is simple. HD for root is 0, a right edge (edge connecting to right subtree) is considered as +1 horizontal distance and a left edge is considered as -1 horizontal distance. For example, in the above tree, HD for Node 4 is at -2, HD for Node 2 is -1, HD for 5 and 6 is 0 and HD for node 7 is +2.
We can do preorder traversal of the given Binary Tree. While traversing the tree, we can recursively calculate HDs. We initially pass the horizontal distance as 0 for root. For left subtree, we pass the Horizontal Distance as Horizontal distance of root minus 1. For right subtree, we pass the Horizontal Distance as Horizontal Distance of root plus 1. For every HD value, we maintain a list of nodes in a hash map. Whenever we see a node in traversal, we go to the hash map entry and add the node to the hash map using HD as a key in a map.

Following is the C++ implementation of the above method. Thanks to Chirag for providing the below C++ implementation.

## C++

 `// C++ program for printing vertical order of a given binary tree ` `#include ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Structure for a binary tree node ` `struct` `Node ` `{ ` `    ``int` `key; ` `    ``Node *left, *right; ` `}; ` ` `  `// A utility function to create a new node ` `struct` `Node* newNode(``int` `key) ` `{ ` `    ``struct` `Node* node = ``new` `Node; ` `    ``node->key = key; ` `    ``node->left = node->right = NULL; ` `    ``return` `node; ` `} ` ` `  `// Utility function to store vertical order in map 'm' ` `// 'hd' is horigontal distance of current node from root. ` `// 'hd' is initially passed as 0 ` `void` `getVerticalOrder(Node* root, ``int` `hd, map<``int``, vector<``int``>> &m) ` `{ ` `    ``// Base case ` `    ``if` `(root == NULL) ` `        ``return``; ` ` `  `    ``// Store current node in map 'm' ` `    ``m[hd].push_back(root->key); ` ` `  `    ``// Store nodes in left subtree ` `    ``getVerticalOrder(root->left, hd-1, m); ` ` `  `    ``// Store nodes in right subtree ` `    ``getVerticalOrder(root->right, hd+1, m); ` `} ` ` `  `// The main function to print vertical order of a binary tree ` `// with the given root ` `void` `printVerticalOrder(Node* root) ` `{ ` `    ``// Create a map and store vertical order in map using ` `    ``// function getVerticalOrder() ` `    ``map < ``int``,vector<``int``> > m; ` `    ``int` `hd = 0; ` `    ``getVerticalOrder(root, hd,m); ` ` `  `    ``// Traverse the map and print nodes at every horigontal ` `    ``// distance (hd) ` `    ``map< ``int``,vector<``int``> > :: iterator it; ` `    ``for` `(it=m.begin(); it!=m.end(); it++) ` `    ``{ ` `        ``for` `(``int` `i=0; isecond.size(); ++i) ` `            ``cout << it->second[i] << ``" "``; ` `        ``cout << endl; ` `    ``} ` `} ` ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `    ``Node *root = newNode(1); ` `    ``root->left = newNode(2); ` `    ``root->right = newNode(3); ` `    ``root->left->left = newNode(4); ` `    ``root->left->right = newNode(5); ` `    ``root->right->left = newNode(6); ` `    ``root->right->right = newNode(7); ` `    ``root->right->left->right = newNode(8); ` `    ``root->right->right->right = newNode(9); ` `    ``cout << ``"Vertical order traversal is n"``; ` `    ``printVerticalOrder(root); ` `    ``return` `0; ` `} `

## Java

 `// Java program for printing vertical order of a given binary tree ` `import` `java.util.TreeMap; ` `import` `java.util.Vector; ` `import` `java.util.Map.Entry; ` ` `  `public` `class` `VerticalOrderBtree  ` `{ ` `    ``// Tree node ` `    ``static` `class` `Node ` `    ``{ ` `        ``int` `key; ` `        ``Node left; ` `        ``Node right; ` `         `  `        ``// Constructor ` `        ``Node(``int` `data) ` `        ``{ ` `            ``key = data; ` `            ``left = ``null``; ` `            ``right = ``null``; ` `        ``} ` `    ``} ` `     `  `    ``// Utility function to store vertical order in map 'm' ` `    ``// 'hd' is horizontal distance of current node from root. ` `    ``// 'hd' is initially passed as 0 ` `    ``static` `void` `getVerticalOrder(Node root, ``int` `hd, ` `                                ``TreeMap> m) ` `    ``{ ` `        ``// Base case ` `        ``if``(root == ``null``) ` `            ``return``; ` `         `  `        ``//get the vector list at 'hd' ` `        ``Vector get =  m.get(hd); ` `         `  `        ``// Store current node in map 'm' ` `        ``if``(get == ``null``) ` `        ``{ ` `            ``get = ``new` `Vector<>(); ` `            ``get.add(root.key); ` `        ``} ` `        ``else` `            ``get.add(root.key); ` `         `  `        ``m.put(hd, get); ` `         `  `        ``// Store nodes in left subtree ` `        ``getVerticalOrder(root.left, hd-``1``, m); ` `         `  `        ``// Store nodes in right subtree ` `        ``getVerticalOrder(root.right, hd+``1``, m); ` `    ``} ` `     `  `    ``// The main function to print vertical order of a binary tree ` `    ``// with the given root ` `    ``static` `void` `printVerticalOrder(Node root) ` `    ``{ ` `        ``// Create a map and store vertical order in map using ` `        ``// function getVerticalOrder() ` `        ``TreeMap> m = ``new` `TreeMap<>(); ` `        ``int` `hd =``0``; ` `        ``getVerticalOrder(root,hd,m); ` `         `  `        ``// Traverse the map and print nodes at every horigontal ` `        ``// distance (hd) ` `        ``for` `(Entry> entry : m.entrySet()) ` `        ``{ ` `            ``System.out.println(entry.getValue()); ` `        ``} ` `    ``} ` `     `  `    ``// Driver program to test above functions ` `    ``public` `static` `void` `main(String[] args) { ` ` `  `        ``// TO DO Auto-generated method stub ` `        ``Node root = ``new` `Node(``1``); ` `        ``root.left = ``new` `Node(``2``); ` `        ``root.right = ``new` `Node(``3``); ` `        ``root.left.left = ``new` `Node(``4``); ` `        ``root.left.right = ``new` `Node(``5``); ` `        ``root.right.left = ``new` `Node(``6``); ` `        ``root.right.right = ``new` `Node(``7``); ` `        ``root.right.left.right = ``new` `Node(``8``); ` `        ``root.right.right.right = ``new` `Node(``9``); ` `        ``System.out.println(``"Vertical Order traversal is"``); ` `        ``printVerticalOrder(root); ` `    ``} ` `} ` `// This code is contributed by Sumit Ghosh `

## Python

 `# Python program for printing vertical order of a given ` `# binary tree ` ` `  `# A binary tree node ` `class` `Node: ` `    ``# Constructor to create a new node ` `    ``def` `__init__(``self``, key): ` `        ``self``.key ``=` `key ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  `# Utility function to store vertical order in map 'm'  ` `# 'hd' is horizontal distance of current node from root ` `# 'hd' is initially passed as 0 ` `def` `getVerticalOrder(root, hd, m): ` ` `  `    ``# Base Case ` `    ``if` `root ``is` `None``: ` `        ``return` `     `  `    ``# Store current node in map 'm' ` `    ``try``: ` `        ``m[hd].append(root.key) ` `    ``except``: ` `        ``m[hd] ``=` `[root.key] ` `     `  `    ``# Store nodes in left subtree ` `    ``getVerticalOrder(root.left, hd``-``1``, m) ` `     `  `    ``# Store nodes in right subtree ` `    ``getVerticalOrder(root.right, hd``+``1``, m) ` ` `  `# The main function to print vertical order of a binary ` `#tree ith given root ` `def` `printVerticalOrder(root): ` `     `  `    ``# Create a map and store vertical order in map using ` `    ``# function getVerticalORder() ` `    ``m ``=` `dict``() ` `    ``hd ``=` `0`  `    ``getVerticalOrder(root, hd, m) ` `     `  `    ``# Traverse the map and print nodes at every horizontal ` `    ``# distance (hd) ` `    ``for` `index, value ``in` `enumerate``(``sorted``(m)): ` `        ``for` `i ``in` `m[value]: ` `            ``print` `i, ` `        ``print`  ` `  ` `  `# Driver program to test above function ` `root ``=` `Node(``1``) ` `root.left ``=` `Node(``2``) ` `root.right ``=` `Node(``3``) ` `root.left.left ``=` `Node(``4``) ` `root.left.right ``=` `Node(``5``) ` `root.right.left ``=` `Node(``6``) ` `root.right.right ``=` `Node(``7``) ` `root.right.left.right ``=` `Node(``8``) ` `root.right.right.right ``=` `Node(``9``) ` `print` `"Vertical order traversal is"` `printVerticalOrder(root) ` ` `  `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) `

Output:

```Vertical order traversal is
4
2
1 5 6
3 8
7
9```

Time Complexity of hashing based solution can be considered as O(n) under the assumption that we have good hashing function that allows insertion and retrieval operations in O(1) time. In the above C++ implementation, map of STL is used. map in STL is typically implemented using a Self-Balancing Binary Search Tree where all operations take O(Logn) time. Therefore time complexity of the above implementation is O(nLogn).

Note that the above solution may print nodes in same vertical order as they appear in tree. For example, the above program prints 12 before 9. See this for a sample run.

```             1
/
2       3
/      /
4    5  6    7
/
8 10  9

11

12      ```

Refer below post for level order traversal based solution. The below post makes sure that nodes of a vertical line are printed in the same order as they appear in the tree.

Print a Binary Tree in Vertical Order | Set 3 (Using Level Order Traversal)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Article Tags :

44

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.