Given a binary tree, print it vertically. The following examples illustrate the vertical order traversal.
Examples:
Input: 1
/ \
2 3
/ \ / \
4 5 6 7
/ \
8 9Output:
4
2
1 5 6
3 8
7
9
Vertical order traversal of the binary tree using Self Balancing BSTs:
To solve the problem follow the below idea:
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.
Follow the below steps to solve the problem:
- Do a 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 the root.
- For the left subtree, we pass the Horizontal Distance as the Horizontal distance of the root minus 1.
- For the right subtree, we pass the Horizontal Distance as the Horizontal Distance of the 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.
Below is the implementation of the above approach, thanks to Chirag for providing the below C++ implementation:
C++
// C++ program for printing vertical order of a given binary// tree#include <bits/stdc++.h>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<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 rootvoid 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 horizontal // 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 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 is \n"; // Function call printVerticalOrder(root); return 0;} |
Java
// Java program for printing vertical order of a given// binary treeimport java.util.Map.Entry;import java.util.TreeMap;import java.util.Vector;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 // horizontal distance (hd) for (Entry<Integer, Vector<Integer> > entry : m.entrySet()) { System.out.println(entry.getValue()); } } // Driver code 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 |
Python3
# Python3 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, end=" ") print()# Driver program to test above functionif __name__ == '__main__': 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) |
C#
// C# program for printing vertical order of a given binary// treeusing System;using System.Collections.Generic;public class VerticalOrderBtree { // Tree node public class Node { public int key; public Node left; public Node right; // Constructor public 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, SortedDictionary<int, List<int> > m) { // Base case if (root == null) return; // get the vector list at 'hd' List<int> get = new List<int>(); // m[hd]; if (m.ContainsKey(hd)) get.AddRange(m[hd]); // Store current node in map 'm' if (get == null) { get = new List<int>(); get.Add(root.key); } else get.Add(root.key); m[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() SortedDictionary<int, List<int> > m = new SortedDictionary<int, List<int> >(); int hd = 0; getVerticalOrder(root, hd, m); // Traverse the map and print nodes at every // horizontal distance (hd) foreach(KeyValuePair<int, List<int> > entry in m) { foreach(int v in entry.Value) Console.Write(v + " "); Console.WriteLine(); } } // Driver code 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); Console.WriteLine("Vertical Order traversal is"); printVerticalOrder(root); }}// This code is contributed by Rajput-Ji |
Javascript
// javascript program for printing vertical order of a given binary// tree// Structure for a binary tree nodeclass Node { constructor(val){ this.key = val; this.left = null; this.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 0function getVerticalOrder(root, hd, m){ // Base case if (root == null) return; // Store current node in map 'm' if((hd in m) == false){ m[hd] = new Array(); } m[hd].push(root.key); // Store nodes in left subtree getVerticalOrder(root.left, hd - 1, m); // Store nodes in right subtree getVerticalOrder(root.right, hd + 1, m);}// custom sort function. function sortFunction(a, b) { if (a[0] === b[0]) { return 0; } else { return (a[0] < b[0]) ? -1 : 1; }}// The main function to print vertical order of a binary// tree with the given rootfunction printVerticalOrder(root){ // Create a map and store vertical order in map using // function getVerticalOrder() let m = {}; let hd = 0; getVerticalOrder(root, hd, m); // Traverse the map and print nodes at every horizontal // distance (hd) // store the map in the array, as keys act as characters, and cannot be sorted numerically. let x = new Array(); for (const key in m) { let y = new Array(); y.push(parseInt(key)); for(let i = 0; i < m[key].length; i++){ y.push(m[key][i]); } x.push(y); } // sort in ascending order. x.sort(sortFunction); // print then. for(let i = 0; i < x.length; i++){ for(let j = 0; j < x[i].length; j++){ if(j == 0) continue; document.write(x[i][j] + " "); } document.write("\n"); } }// Driver codelet 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);console.log("Vertical order traversal is \n");// Function callprintVerticalOrder(root);// The code is contributed by Nidhi goel. |
Output
Vertical order traversal is 4 2 1 5 6 3 8 7 9
Time Complexity: O(N log N). The hashing based solution can be considered as O(N) under the assumption that we have a 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(Log N) time.
Auxiliary Space: O(N)
Note: The above solution may not print nodes in the same vertical order as they appear in the tree.
For example, the above program prints 12 before 9. See this for a sample run.
1
/ \
2 3
/ \ / \
4 5 6 7
/ \
8 9
/ \
10 11
\
12
Refer below post for a 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)
Print vertical order traversal of the binary tree in the same order as they appear:
To solve the problem follow the below idea:
We can also maintain the order of nodes in the same vertical order as they appear in the tree. Nodes having the same horizontal distance will print according to level order.
For example, In below diagram 9 and 12 have the same horizontal distance. We can make sure that if a node like 12 comes below in the same vertical line, it is printed after a node like 9Idea: Instead of using horizontal distance as a key in the map, we will use horizontal distance + vertical distance as the key. We know that the number of nodes can’t be more than the integer range in a binary tree.
We will use the first 30 bits of the key for horizontal distance [MSB to LSB] and will use the 30 next bits for vertical distance. Thus keys will be stored in the map as per our requirement.
Follow the below steps to solve the problem:
- Declare a map to store the value of nodes at each level
- If the root is null then return from the function(Base case)
- Create an integer val and set its value to horizontal distance << 30 OR vertical distance
- Push root->data in the map using val as the key
- Recur for root->left and root->right with horizontal distance – 1, vertical distance + 1 and horizontal distance + 1, vertical distance -1 respectively
- Print the solution using map
Below is the implementation of the above approach:
C++14
// 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 distancevoid 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 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;} |
Java
// Java program for printing vertical order of a given// binary treeimport java.util.Map.Entry;import java.util.TreeMap;import java.util.Vector;public class VerticalOrderBtree { // Tree node static class Node { int data; Node left; Node right; // Constructor Node(int key) { data = key; left = null; right = null; } } // Store vertical order // in map "m", hd = horizontal // distance, vd = vertical distance static void preOrderTraversal(Node root, long hd, long vd, TreeMap<Long, Vector<Integer> > m) { if (root == null) 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 val = hd << 30 | vd; // insert in map if (m.get(val) != null) m.get(val).add(root.data); else { Vector<Integer> v = new Vector<Integer>(); v.add(root.data); m.put(val, v); } preOrderTraversal(root.left, hd - 1, vd + 1, m); preOrderTraversal(root.right, hd + 1, vd + 1, m); } static void verticalOrder(Node root) { // map to store all nodes in vertical order. // keys will be horizontal + vertical distance. TreeMap<Long, Vector<Integer> > mp = new TreeMap<>(); preOrderTraversal(root, 0, 1, mp); // print map int prekey = Integer.MAX_VALUE; for (Entry<Long, Vector<Integer> > entry : mp.entrySet()) { if (prekey != Integer.MAX_VALUE && (entry.getKey() >> 30) != prekey) { System.out.println(); } prekey = (int)(entry.getKey() >> 30); for (int x : entry.getValue()) System.out.print(x + " "); } } // Driver code public static void main(String[] args) { 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 :- "); verticalOrder(root); }}// This code is contributed by Abhijeet Kumar(abhijeet19403) |
Python3
import sys# Python program for printing# vertical order of a given binary# treeclass Node: def __init__(self, data): self.data = data self.left = None self.right = None# Store vertical order# in map "m", hd = horizontal# distance, vd = vertical distancedef preOrderTraversal(root, hd, vd, m): if not 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. val = hd << 30 | vd # insert in map if val in m: m[val].append(root.data) else: m[val] = [root.data] preOrderTraversal(root.left, hd-1, vd+1, m) preOrderTraversal(root.right, hd+1, vd+1, m)def verticalOrder(root): mp = dict() preOrderTraversal(root, 0, 0, mp) # print dictionary prekey = sys.maxsize for i in sorted(mp.keys()): if (prekey != sys.maxsize) and (i >> 30 != prekey): print() prekey = i >> 30 for j in mp[i]: print(j, end=" ")# Driver coderoot = 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 :- ")verticalOrder(root)# This code is contributed by prashantchandelme. |
C#
// C# program for printing vertical order of a given binary// treeusing System;using System.Collections.Generic;public class VerticalOrderBtree { // Tree node public class Node { public int data; public Node left; public Node right; // Constructor public Node(int val) { data = val; left = null; right = null; } } // Store vertical order // in map "m", hd = horizontal // distance, vd = vertical distance static void preOrderTraversal(Node root, long hd, long vd, SortedDictionary<long, List<int> > m) { if (root == null) 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 val = hd << 30 | vd; // insert in map if (m.ContainsKey(val) == true) m[val].Add(root.data); else { List<int> v = new List<int>(); v.Add(root.data); m.Add(val, v); } preOrderTraversal(root.left, hd - 1, vd + 1, m); preOrderTraversal(root.right, hd + 1, vd + 1, m); } static void verticalOrder(Node root) { // map to store all nodes in vertical order. // keys will be horizontal + vertical distance. SortedDictionary<long, List<int> > mp = new SortedDictionary<long, List<int> >(); preOrderTraversal(root, 0, 1, mp); // print map int prekey = Int32.MaxValue; foreach(KeyValuePair<long, List<int> > entry in mp) { if (prekey != Int32.MaxValue && (entry.Key >> 30) != prekey) { Console.WriteLine(); } prekey = (int)entry.Key >> 30; foreach(int v in entry.Value) Console.Write(v + " "); } } // Driver program to test above functions public static void Main(String[] args) { 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); Console.WriteLine("Vertical Order traversal :- "); verticalOrder(root); }}// This code is contributed by Abhijeet Kumar(abhijeet19403) |
Javascript
class Node { constructor(data) { this.data = data; this.left = null; this.right = null; }}function preOrderTraversal(root, hd, vd, 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. const val = BigInt(hd << 30n | vd); // insert in map if (!m.has(val)) { m.set(val, []); } m.get(val).push(root.data); preOrderTraversal(root.left, hd - 1n, vd + 1n, m); preOrderTraversal(root.right, hd + 1n, vd + 1n, m);}function verticalOrder(root) { // map to store all nodes in vertical order. // keys will be horizontal + vertical distance. const mp = new Map(); preOrderTraversal(root, 0n, 1n, mp); // print map let prekey = BigInt(Number.MAX_SAFE_INTEGER); for (const [key, value] of [...mp.entries()].sort((a, b) => { // sort map entries by key if (a[0] < b[0]) return -1; if (a[0] > b[0]) return 1; return 0; })) { if (prekey !== (key >> 30n)) { console.log(); } prekey = key >> 30n; process.stdout.write(value.join(" ") + " "); }}// Driver codeconst 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);process.stdout.write("Vertical order traversal :- ");verticalOrder(root); |
Output
Vertical order traversal :- 4 2 1 5 6 3 8 7 9
Time Complexity: O(N Log N)
Auxiliary Space: O(N)
Vertical order traversal of the binary tree using computeIfAbsent method in Java:
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 the above approach:
C++
#include <bits/stdc++.h>using namespace std;// structure of the tree nodeclass Node {public: int data; Node* left; Node* right; Node(int item) { data = item; left = right = NULL; }};class BinaryTree {public: Node* root; // Values class class Values { public: int max, min; }; // Program to find vertical Order void verticalOrder(Node* node) { Values val; // Create TreeMap map<int, vector<int> > mp; // Function Call to findHorizontalDistance findHorizontalDistance(node, &val, &val, 0, mp); // Iterate over map.values() for (auto list : mp) { for (auto element : list.second) { cout << element << " "; } cout << endl; } } // Program to find Horizontal Distance void findHorizontalDistance(Node* node, Values* min, Values* max, int hd, map<int, vector<int> >& mp) { // 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 mp[hd].push_back(node->data); // Function Call with hd equal to hd - 1 findHorizontalDistance(node->left, min, max, hd - 1, mp); // Function Call with hd equal to hd + 1 findHorizontalDistance(node->right, min, max, hd + 1, mp); }};int main(){ BinaryTree tree; // 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); cout << "vertical order traversal is :\n"; // Function Call tree.verticalOrder(tree.root); return 0;} |
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<Integer, List<Integer> > map = new TreeMap<Integer, List<Integer> >(); // Function Call to findHorizontalDistance findHorizontalDistance(node, val, val, 0, map); // Iterate over map.values() for (List<Integer> list : map.values()) { System.out.println(list); } } // Program to find Horizontal Distance public void findHorizontalDistance(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 findHorizontalDistance(node.left, min, max, hd - 1, map); // Function Call with hd equal to hd + 1 findHorizontalDistance(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); }} |
C#
// C# Program for above approachusing System;using System.Collections.Generic;public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}public class BinaryTree { public Node root; // Values class public class Values { public int max, min; } // Program to find vertical Order public void verticalOrder(Node node) { Values val = new Values(); // Create TreeMap SortedDictionary<int, List<int> > map = new SortedDictionary<int, List<int> >(); // Function Call to findHorizontalDistance findHorizontalDistance(node, val, val, 0, map); // Iterate over map.values() foreach(List<int> list in map.Values) { Console.WriteLine(String.Join(" ", list)); } } // Program to find Horizontal Distance public static void findHorizontalDistance( Node node, Values min, Values max, int hd, SortedDictionary<int, List<int> > 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 TryGetValue and Add methods if (!map.TryGetValue(hd, out List<int> value)) { value = new List<int>(); map.Add(hd, value); } value.Add(node.data); // Function Call with hd equal to hd - 1 findHorizontalDistance(node.left, min, max, hd - 1, map); // Function Call with hd equal to hd + 1 findHorizontalDistance(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); Console.WriteLine("vertical order traversal is :"); // Function Call tree.verticalOrder(tree.root); }}// The code is contributed by Arushi Goel. |
Python3
class Node: def __init__(self, item): self.data = item self.left = None self.right = Noneclass BinaryTree: def __init__(self): self.root = None class Values: def __init__(self): self.max = float('-inf') self.min = float('inf') def verticalOrder(self, node): val = self.Values() # Create dictionary to store vertical order mp = {} # Function call to findHorizontalDistance self.findHorizontalDistance(node, val, val, 0, mp) # Iterate over dictionary values for hd, lst in sorted(mp.items()): for element in lst: print(element, end=" ") print() def findHorizontalDistance(self, node, minVal, maxVal, hd, mp): # If node is None if node is None: return # if hd is less than min.min if hd < minVal.min: minVal.min = hd # if hd is greater than min.min if hd > maxVal.max: maxVal.max = hd # Using setdefault mp.setdefault(hd, []).append(node.data) # Function Call with hd equal to hd - 1 self.findHorizontalDistance(node.left, minVal, maxVal, hd - 1, mp) # Function Call with hd equal to hd + 1 self.findHorizontalDistance(node.right, minVal, maxVal, hd + 1, mp)if __name__ == '__main__': tree = BinaryTree() # Let us construct the tree shown in above diagram tree.root = Node(1) tree.root.left = Node(2) tree.root.right = Node(3) tree.root.left.left = Node(4) tree.root.left.right = Node(5) tree.root.right.left = Node(6) tree.root.right.right = Node(7) tree.root.right.left.right = Node(8) tree.root.right.right.right = Node(9) print("Vertical order traversal is:\n") # Function Call tree.verticalOrder(tree.root) |
Javascript
// Javascript code addition class Node { constructor(item) { this.data = item; this.left = null; this.right = null; }}class BinaryTree { constructor() { this.root = null; } verticalOrder(node) { let val = new Values(); // Create dictionary to store vertical order let mp = {}; // Function call to findHorizontalDistance this.findHorizontalDistance(node, val, val, 0, mp); // Iterate over dictionary values for (let hd of Object.keys(mp).sort((a, b) => a - b)) { let lst = mp[hd]; for (let element of lst) { process.stdout.write(element + " "); } process.stdout.write("\n"); } } findHorizontalDistance(node, minVal, maxVal, hd, mp) { // If node is null if (!node) { return; } // if hd is less than min.min if (hd < minVal.min) { minVal.min = hd; } // if hd is greater than min.min if (hd > maxVal.max) { maxVal.max = hd; } // Using setdefault if (!mp[hd]) { mp[hd] = []; } mp[hd].push(node.data); // Function Call with hd equal to hd - 1 this.findHorizontalDistance(node.left, minVal, maxVal, hd - 1, mp); // Function Call with hd equal to hd + 1 this.findHorizontalDistance(node.right, minVal, maxVal, hd + 1, mp); }}class Values { constructor() { this.max = -Infinity; this.min = Infinity; }}let tree = new BinaryTree();// Let us construct the tree shown in above diagramtree.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);process.stdout.write("Vertical order traversal is:\n");// Function Calltree.verticalOrder(tree.root);// The code is contributed by Arushi Goel. |
Output
vertical order traversal is : [4] [2] [1, 5, 6] [3, 8] [7] [9]
Time Complexity: O(N Log N)
Auxiliary Space: O(N)
Vertical order traversal of the binary tree using the Unordered Map method:
Note: We have seen the ordered map above but, its complexity is O(N log N), and also it does not print the vertical nodes of the same horizontal distance in the correct order.
Here we implement this using an unordered map, as the unordered map is implemented using a hash table its complexity is O(n), better than using an ordered map which is implemented using a BST.
Follow the below steps to solve the problem:
- Create a queue of pair to store the node and its horizontal distance in the tree
- Create a map to store the value of nodes at each horizontal distance
- Now perform a BFS on the tree
- At each iteration store the nodes with a particular horizontal distance in the map
- Push the left and the right child of the tree with horizontal distance – 1 and horizontal distance + 1 into the queue
- Print the answer using map
Note: Here for printing all nodes of the same horizontal distance from the root we use mn and mx two variables that store the minimum and maximum horizontal distance from the root:
C++
// C++ program for printing vertical// order of a given binary tree using BFS#include <bits/stdc++.h>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<int, vector<int> > m; int hd = 0; // Create queue to do level order // traversal Every item of queue contains // node and horizontal distance queue<pair<Node*, int> > 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<Node*, int> 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 < hd) mx = hd; } // run the loop from minimum to maximum // every horizontal distance hd for (int i = mn; i <= mx; i++) { vector<int> tmp = m[i]; for (int j = 0; j < tmp.size(); j++) cout << tmp[j] << " "; cout << endl; }}// 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 is \n"; printVerticalOrder(root); return 0;} |
Java
// Java program for printing vertical// order of a given binary tree using BFSimport java.io.*;import java.util.*;class GFG { // Structure for a binary tree node static class Node { int key; Node left, right; Node(int key) { this.key = key; left = null; right = null; } } static class pair { Node first; int second; pair(Node first, int second) { this.first = first; this.second = second; } } // The main function to print vertical // order of a binary tree with given root static void printVerticalOrder(Node root) { // Base case if (root == null) return; // Create a map and store vertical // order in map using // function getVerticalOrder() HashMap<Integer, ArrayList<Integer> > m = new HashMap<>(); int hd = 0; // Create queue to do level order // traversal Every item of queue contains // node and horizontal distance Queue<pair> q = new ArrayDeque<>(); q.add(new pair(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.remove(); hd = temp.second; Node node = temp.first; // insert this node's data // in vector of hash if (!m.containsKey(hd)) m.put(hd, new ArrayList<>()); m.get(hd).add(node.key); if (node.left != null) q.add(new pair(node.left, hd - 1)); if (node.right != null) q.add(new pair(node.right, hd + 1)); // Update mn and mx if (mn > hd) mn = hd; else if (mx < hd) mx = hd; } // run the loop from minimum to maximum // every horizontal distance hd for (int i = mn; i <= mx; i++) { ArrayList<Integer> tmp = m.get(i); for (int j = 0; j < tmp.size(); j++) System.out.print(tmp.get(j) + " "); System.out.println(); } } // Driver code public static void main(String[] args) { 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 karandeep1234 |
C#
// C# program for printing vertical// order of a given binary tree using BFSusing System;using System.Collections.Generic;// Driver codeclass GFG{ // Structure for a binary tree node class Node { public int key; public Node left, right; public Node (int key) { this.key = key; left = null; right = null; } } class pair { public Node first; public int second; public pair (Node first, int second) { this.first = first; this.second = second; } } // The main function to print vertical // order of a binary tree with given root static void printVerticalOrder (Node root) { // Base case if (root == null) return; // Create a map and store vertical // order in map using // function getVerticalOrder() Dictionary < int, List < int >>m = new Dictionary < int, List < int >>(); int hd = 0; // Create queue to do level order // traversal Every item of queue contains // node and horizontal distance Queue < pair > q = new Queue < pair > (); q.Enqueue (new pair (root, hd)); // mn and mx contain the minimum and // maximum horizontal distance from root int mn = 0, mx = 0; while (q.Count > 0) { // pop from queue front pair temp = q.Dequeue (); hd = temp.second; Node node = temp.first; // insert this node's data // in vector of hash if (!m.ContainsKey (hd)) m.Add (hd, new List < int >()); m[hd].Add (node.key); if (node.left != null) q.Enqueue (new pair (node.left, hd - 1)); if (node.right != null) q.Enqueue (new pair (node.right, hd + 1)); // Update mn and mx if (mn > hd) mn = hd; else if (mx < hd) mx = hd; } // run the loop from minimum to maximum // every horizontal distance hd for (int i = mn; i <= mx; i++) { List < int >tmp = m[i]; for (int j = 0; j < tmp.Count; j++) Console.Write (tmp[j] + " "); Console.WriteLine (); } } static void Main () { 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); Console.WriteLine ("Vertical order traversal is "); printVerticalOrder (root); }} |
Javascript
// Javascript program for printing vertical// order of a given binary tree using BFS // program to implement queue data structure class Queue { constructor() { this.items = Array.from(Array(), () => new Array()); } // add element to the queue push(element) { return this.items.push(element); } // remove element from the queue pop() { if (this.items.length > 0) { return this.items.shift(); } } // view the first element front() { return this.items[0]; } // check if the queue is empty empty() { return this.items.length == 0; } // the size of the queue size() { return this.items.length; } } // Structure for a binary tree node class Node { constructor(key) { this.key = key; this.left = null; this.right = null; } } // The main function to print vertical // order of a binary tree with given root function printVerticalOrder(root) { // Base case if (!root) return; // Create a map and store vertical // order in map using // function getVerticalOrder() let m = new Map(); let hd = 0; // Create queue to do level order // traversal Every item of queue contains // node and horizontal distance let q = new Queue(); q.push([root, hd]); // mn and mx contain the minimum and // maximum horizontal distance from root let mn = 0, mx = 0; while (q.size() > 0) { // pop from queue front temp = q.front(); q.pop(); hd = temp[1]; node = temp[0]; //insert this node's data //in string of hash if (m.get(hd) === undefined) { m.set(hd, "" + node.key); } else { m.set(hd, m.get(hd) + " " + 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 < hd) mx = hd; } // run the loop from minimum to maximum // every horizontal distance hd for (let i = mn; i <= mx; i++) { tmp = m.get(i); console.log(tmp); } } // Driver code let 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); console.log("Vertical order traversal is "); printVerticalOrder(root); |
Python3
# Python program for printing vertical order of a given binary tree using BFS# Structure for a binary tree nodeclass Node: def __init__(self, key): self.key = key self.left = None self.right = None# A function to create a new nodedef newNode(key): node = Node(key) return node# The main function to print vertical# order of a binary tree with given rootdef printVerticalOrder(root): # Base case if not root: return # Create a dictionary and store vertical # order in dictionary using # function getVerticalOrder() m = {} hd = 0 # Create queue to do level order # traversal Every item of queue contains # node and horizontal distance q = [] q.append((root, hd)) # mn and mx contain the minimum and # maximum horizontal distance from root mn, mx = 0, 0 while q: # pop from queue front temp = q.pop(0) hd = temp[1] node = temp[0] # insert this node's data # in vector of hash if hd in m: m[hd].append(node.key) else: m[hd] = [node.key] if node.left: q.append((node.left, hd - 1)) if node.right: q.append((node.right, hd + 1)) # Update mn and mx if mn > hd: mn = hd elif mx < hd: mx = hd # run the loop from minimum to maximum # every horizontal distance hd for i in range(mn, mx+1): if i in m: tmp = m[i] for j in range(len(tmp)): print(tmp[j], end=' ') print()# Driver codeif __name__ == '__main__': 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) print("Vertical order traversal is") printVerticalOrder(root) |
Output
Vertical order traversal is 4 2 1 5 6 3 8 7 9
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
