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

• Difficulty Level : Medium
• Last Updated : 24 Oct, 2021

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

Become a success story instead of just reading about them. Prepare for coding interviews at Amazon and other top product-based companies with our Amazon Test Series. Includes topic-wise practice questions on all important DSA topics along with 10 practice contests of 2 hours each. Designed by industry experts that will surely help you practice and sharpen your programming skills. Wait no more, start your preparation today!

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 nodestruct Node{    int key;    Node *left, *right;}; // A utility function to create a new nodestruct 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 horizontal distance of current node from root.// 'hd' is initially passed as 0void getVerticalOrder(Node* root, int hd, map> &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 rootvoid printVerticalOrder(Node* root){    // Create a map and store vertical order in map using    // function getVerticalOrder()    map < int,vector > m;    int hd = 0;    getVerticalOrder(root, hd,m);     // Traverse the map and print nodes at every horizontal    // distance (hd)    map< int,vector > :: 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 functionsint 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 treeimport 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 horizontal        // 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 nodeclass 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 0def 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 rootdef 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 functionroot = 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 n4
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 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

 // C++ program for printing// vertical order of a given binary// tree#include 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 distancevoid preOrderTraversal(Node* root,                       long long int hd,                       long long int vd,                       map >& 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 > mp;     preOrderTraversal(root, 0, 1, mp);     // print map    int prekey = INT_MAX;    map >::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 codeint 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;}
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

 // Java Program for above approachimport java.util.ArrayList;import java.util.List;import java.util.Map;import java.util.TreeMap; class Node {    int data;    Node left, right;     Node(int item)    {        data = item;        left = right = null;    }} 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 > map            = new TreeMap >();         // Function Call to findHorizonatalDistance        findHorizonatalDistance(node, val, val, 0, map);         // Iterate over map.values()        for (List 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 > 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())            .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);    }}
Output
vertical order traversal is :
[4]
[2]
[1, 5, 6]
[3, 8]
[7]
[9]
done

Another Approach using Unordered Map method:

we have seen ordered map above but, its complexity is O(n logn), and also it does not print the vertical nodes of same horizontal distance in correct order.

here we implement this using an unordered map, as unordered map is implemented using a hash table its complexity is O(n), better then using ordered map which is implemented using a BST.

here for printing all nodes of of same horizontal distance from root we use mn and mx two variables which store the minimum and maximum horizontal distance from root.

C++

 // C++ program for printing vertical// order of a given binary tree using BFS#include  using namespace std; // Structure for a binary tree nodestruct Node {    int key;    Node *left, *right;}; // A function to create a new nodeNode* newNode(int key){    Node* node = new Node;    node->key = key;    node->left = node->right = NULL;    return node;} // The main function to print vertical// order of a binary tree with given rootvoid printVerticalOrder(Node* root){    // Base case    if (!root) return;     // Create a map and store vertical    // order in map using      // function getVerticalOrder()    unordered_map > m;    int hd = 0;     // Create queue to do level order    // traversal Every item of queue contains    // node and horizontal distance    queue > q;    q.push({root, hd});       // mn and mx contain the minimum and      // maximum horizontal distance from root    int mn=0,mx=0;    while (q.size()>0) {               // pop from queue front        pair temp = q.front();        q.pop();        hd = temp.second;        Node* node = temp.first;         // insert this node's data          // in vector of hash        m[hd].push_back(node->key);         if (node->left)            q.push({node->left, hd - 1});        if (node->right)            q.push({node->right, hd + 1});               // Update mn and mx        if(mn>hd)          mn=hd;        else if(mx tmp=m[i];        for(int j=0;jleft = 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);    root->right->right->left = newNode(10);    root->right->right->left->right = newNode(11);    root->right->right->left->right->right = newNode(12);    cout << "Vertical order traversal is \n";    printVerticalOrder(root);    return 0;}
Output
Vertical order traversal is
4
2
1 5 6
3 8 10
7 11
9 12

Time complexity: O(n)

space complexity: O(n)

Here all the nodes of same horizontal distance will be printed in correct order i.e from top to bottom.

My Personal Notes arrow_drop_up