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

Given a binary tree, print it vertically. The following example illustrates the 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

 

print-binary-tree-in-vertical-order

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++

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// C++ program for printing vertical order of a given binary tree
#include <iostream>
#include <vector>
#include <map>
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; i<it->second.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;
}

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Java

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// 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<Integer,Vector<Integer>> m)
    {
        // Base case
        if(root == null)
            return;
         
        //get the vector list at 'hd'
        Vector<Integer> 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<Integer,Vector<Integer>> m = new TreeMap<>();
        int hd =0;
        getVerticalOrder(root,hd,m);
         
        // Traverse the map and print nodes at every horigontal
        // distance (hd)
        for (Entry<Integer, Vector<Integer>> 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

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Python

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# 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)

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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 not 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)

Using Preorder Traversal Approach, Maintain the Order of Nodes in Same Vertical Order as They Appear in Tree: 

We can also maintain the order of nodes in same vertical order as they appear in the tree. Nodes having same horizontal distance will print according to level order.  

For example, In below diagram 9 and 12 have same horizontal distance. We can make sure that  if a node like 12 comes below in same vertical line, it is printed after a node like 9

Idea: Instead of using horizontal distance as a key in the map, we will use  horizontal distance + vertical distance as key. We know that the number of nodes can’t be more than integer range in a binary tree.

We will use first 30 bits of key for horizontal distance [MSB to LSB] and will use 30 next bits for vertical distance. Thus keys will be stored in map as per our requirement.

Below is the implementation of above approach.

C++14

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// C++ program for printing
// vertical order of a given binary
// tree
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int data;
    Node *left, *right;
};
 
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return node;
}
 
// Store vertical order
// in map "m", hd = horizontal
// distance, vd = vertical distance
void preOrderTraversal(Node* root,
                       long long int hd,
                       long long int vd,
                       map<long long int,
                       vector<int> >& m)
{
    if (!root)
        return;
    // key = horizontal
    // distance (30 bits) + vertical
    // distance (30 bits) map
    // will store key in sorted
    // order. Thus nodes having same
    // horizontal distance
    // will sort according to
    // vertical distance.
    long long val = hd << 30 | vd;
 
    // insert in map
    m[val].push_back(root->data);
 
    preOrderTraversal(root->left, hd - 1, vd + 1, m);
    preOrderTraversal(root->right, hd + 1, vd + 1, m);
}
 
void verticalOrder(Node* root)
{
    // map to store all nodes in vertical order.
    // keys will be horizontal + vertical distance.
    map<long long int, vector<int> > mp;
 
    preOrderTraversal(root, 0, 1, mp);
 
    // print map
    int prekey = INT_MAX;
    map<long long int, vector<int> >::iterator it;
    for (it = mp.begin(); it != mp.end(); it++) {
        if (prekey != INT_MAX
            && (it->first >> 30) != prekey) {
            cout << endl;
        }
        prekey = it->first >> 30;
        for (int j = 0; j < it->second.size(); j++)
            cout << it->second[j] << " ";
    }
}
 
// Driver code
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 :- " << endl;
    verticalOrder(root);
    return 0;
}

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Output



Vertical order traversal :- 
4 
2 
1 5 6 
3 8 
7 
9 

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

Auxiliary Space: O(n)

Another Approach using computeIfAbsent method:

We can write the code in a more concise way, by using computeIfAbsent method of the map in java and by using a treemap for natural sorting based upon keys.

Below is the implementation of above approach.

Java

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// Java Program for above approach
import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.TreeMap;
 
public class BinaryTree
{
   
  Node root;
   
  // Values class
  class Values
  {
    int max, min;
  }
 
  // Program to find vertical Order
  public void verticalOrder(Node node)
  {
    Values val = new Values();
     
    // Create TreeMap
    Map<Integer,
        List<Integer>> map = new TreeMap<Integer,
                                   List<Integer>>();
     
    // Function Call to findHorizonatalDistance
    findHorizonatalDistance(node, val, val,
                                         0, map);
     
    // Iterate over map.values()
    for(List<Integer> list : map.values())
    {
      System.out.println(list);
    }
     
    // Print "done"
    System.out.println("done");
  }
   
   
  // Program to find Horizonatal Distance
  public void findHorizonatalDistance(Node node,
                      Values min, Values max, int hd,
                      Map<Integer, List<Integer>> map)
  {
     
    // If node is null
    if(node == null)
      return;
     
    // if hd is less than min.min
    if(hd < min.min)
      min.min=hd;
     
    // if hd is greater than min.min
    if(hd > max.max)
      max.max=hd;
     
    // Using computeIfAbsent
    map.computeIfAbsent(hd,
             k->new ArrayList<Integer>()).
                                 add(node.data);
     
     
    // Function Call with hd equal to hd - 1
    findHorizonatalDistance(node.left, min,
                                 max, hd-1, map);
     
    // Function Call with hd equal to hd + 1
    findHorizonatalDistance(node.right, min,
                                  max, hd+1, map);
  }
   
  
  // Driver Code
  public static void main(String[] args)
  {
     
    BinaryTree tree = new BinaryTree();
     
    /* Let us construct the tree shown
                         in above diagram */
    tree.root = new Node(1);
    tree.root.left = new Node(2);
    tree.root.right = new Node(3);
    tree.root.left.left = new Node(4);
    tree.root.left.right = new Node(5);
    tree.root.right.left = new Node(6);
    tree.root.right.right = new Node(7);
    tree.root.right.left.right = new Node(8);
    tree.root.right.right.right = new Node(9);
 
    System.out.println("vertical order
                             traversal is :");
                        
    // Function Call
    tree.verticalOrder(tree.root);
  }
}

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Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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