Vertical width of Binary tree | Set 1
Last Updated :
11 Mar, 2023
Given a Binary tree, the task is to find the vertical width of the Binary tree. The width of a binary tree is the number of vertical paths in the Binary tree.
Examples:
Input:
Output: 6
Explanation: In this image, the tree contains 6 vertical lines which are the required width of the tree.
Input :
7
/ \
6 5
/ \ / \
4 3 2 1
Output : 5
Input:
1
/ \
2 3
/ \ / \
4 5 6 7
\ \
8 9
Output : 6
Approach:
Take inorder traversal and then take a temp variable to keep the track of unique vertical paths. when moving to left of the node then temp decreases by one and if goes to right then temp value increases by one. If the minimum is greater than temp, then update minimum = temp and if maximum less than temp then update maximum = temp. The vertical width of the tree will be equal to abs(minimum ) + maximum.
Follow the below steps to Implement the idea:
- Initialize a minimum and maximum variable to track the left-most and right-most index.
- Run Depth-first search traversal and maintain the current horizontal index curr, for root curr = 0.
- Traverse left subtree with curr = curr-1
- Update maximum with curr if curr > maximum.
- Update minimum with curr if curr < minimum.
- Traverse right subtree with curr = curr + 1.
- Print the vertical width of the tree i.e. abs(minimum ) + maximum.
Below is the Implementation of the above approach:
C++
// CPP program to print vertical width
// of a tree
#include <bits/stdc++.h>
using namespace std;
// A Binary Tree Node
struct Node {
int data;
struct Node *left, *right;
};
// get vertical width
void lengthUtil(Node* root, int& maximum, int& minimum,
int curr = 0)
{
if (root == NULL)
return;
// traverse left
lengthUtil(root->left, maximum, minimum, curr - 1);
// if curr is decrease then get
// value in minimum
if (minimum > curr)
minimum = curr;
// if curr is increase then get
// value in maximum
if (maximum < curr)
maximum = curr;
// traverse right
lengthUtil(root->right, maximum, minimum, curr + 1);
}
int getLength(Node* root)
{
int maximum = 0, minimum = 0;
lengthUtil(root, maximum, minimum, 0);
// 1 is added to include root in the width
return (abs(minimum) + maximum) + 1;
}
// Utility function to create a new tree node
Node* newNode(int data)
{
Node* curr = new Node;
curr->data = data;
curr->left = curr->right = NULL;
return curr;
}
// Driver program to test above functions
int main()
{
Node* root = newNode(7);
root->left = newNode(6);
root->right = newNode(5);
root->left->left = newNode(4);
root->left->right = newNode(3);
root->right->left = newNode(2);
root->right->right = newNode(1);
cout << getLength(root) << "\n";
return 0;
}
// Java program to print vertical width
// of a tree
import java.util.*;
class GFG {
// A Binary Tree Node
static class Node {
int data;
Node left, right;
};
static int maximum = 0, minimum = 0;
// get vertical width
static void lengthUtil(Node root, int curr)
{
if (root == null)
return;
// traverse left
lengthUtil(root.left, curr - 1);
// if curr is decrease then get
// value in minimum
if (minimum > curr)
minimum = curr;
// if curr is increase then get
// value in maximum
if (maximum < curr)
maximum = curr;
// traverse right
lengthUtil(root.right, curr + 1);
}
static int getLength(Node root)
{
maximum = 0;
minimum = 0;
lengthUtil(root, 0);
// 1 is added to include root in the width
return (Math.abs(minimum) + maximum) + 1;
}
// Utility function to create a new tree node
static Node newNode(int data)
{
Node curr = new Node();
curr.data = data;
curr.left = curr.right = null;
return curr;
}
// Driver Code
public static void main(String[] args)
{
Node root = newNode(7);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(4);
root.left.right = newNode(3);
root.right.left = newNode(2);
root.right.right = newNode(1);
System.out.println(getLength(root));
}
}
// This code is contributed by Rajput-Ji
# Python3 program to print vertical width
# of a tree
# class to create a new tree node
class newNode:
def __init__(self, data):
self.data = data
self.left = self.right = None
# get vertical width
def lengthUtil(root, maximum, minimum, curr=0):
if (root == None):
return
# traverse left
lengthUtil(root.left, maximum,
minimum, curr - 1)
# if curr is decrease then get
# value in minimum
if (minimum[0] > curr):
minimum[0] = curr
# if curr is increase then get
# value in maximum
if (maximum[0] < curr):
maximum[0] = curr
# traverse right
lengthUtil(root.right, maximum,
minimum, curr + 1)
def getLength(root):
maximum = [0]
minimum = [0]
lengthUtil(root, maximum, minimum, 0)
# 1 is added to include root in the width
return (abs(minimum[0]) + maximum[0]) + 1
# Driver Code
if __name__ == '__main__':
root = newNode(7)
root.left = newNode(6)
root.right = newNode(5)
root.left.left = newNode(4)
root.left.right = newNode(3)
root.right.left = newNode(2)
root.right.right = newNode(1)
print(getLength(root))
# This code is contributed by PranchalK
// C# program to print vertical width
// of a tree
using System;
class GFG {
// A Binary Tree Node
public class Node {
public int data;
public Node left, right;
};
static int maximum = 0, minimum = 0;
// get vertical width
static void lengthUtil(Node root, int curr)
{
if (root == null)
return;
// traverse left
lengthUtil(root.left, curr - 1);
// if curr is decrease then get
// value in minimum
if (minimum > curr)
minimum = curr;
// if curr is increase then get
// value in maximum
if (maximum < curr)
maximum = curr;
// traverse right
lengthUtil(root.right, curr + 1);
}
static int getLength(Node root)
{
maximum = 0;
minimum = 0;
lengthUtil(root, 0);
// 1 is added to include root in the width
return (Math.Abs(minimum) + maximum) + 1;
}
// Utility function to create a new tree node
static Node newNode(int data)
{
Node curr = new Node();
curr.data = data;
curr.left = curr.right = null;
return curr;
}
// Driver Code
public static void Main(String[] args)
{
Node root = newNode(7);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(4);
root.left.right = newNode(3);
root.right.left = newNode(2);
root.right.right = newNode(1);
Console.WriteLine(getLength(root));
}
}
// This code is contributed by PrinciRaj1992
<script>
// JavaScript program to prvertical width
// of a tree
// class to create a new tree node
class newNode{
constructor(data){
this.data = data
this.left = this.right = null
}
}
// get vertical width
function lengthUtil(root, maximum, minimum, curr = 0){
if (root == null)
return
// traverse left
lengthUtil(root.left, maximum,
minimum, curr - 1)
// if curr is decrease then get
// value in minimum
if (minimum[0] > curr)
minimum[0] = curr
// if curr is increase then get
// value in maximum
if (maximum[0] <curr)
maximum[0] = curr
// traverse right
lengthUtil(root.right, maximum,
minimum, curr + 1)
}
function getLength(root)
{
let maximum = [0]
let minimum = [0]
lengthUtil(root, maximum, minimum, 0)
// 1 is added to include root in the width
return (Math.abs(minimum[0]) + maximum[0]) + 1
}
// Driver Code
root = new newNode(7)
root.left = new newNode(6)
root.right = new newNode(5)
root.left.left = new newNode(4)
root.left.right = new newNode(3)
root.right.left = new newNode(2)
root.right.right = new newNode(1)
document.write(getLength(root),"</br>")
// This code is contributed by shinjanpatra.
</script>
Time Complexity: O(n)
Auxiliary Space: O(h) where h is the height of the binary tree. This much space is needed for recursive calls.
Below is the idea to solve the problem:
Create a class to store the node and the horizontal level of the node. The horizontal level of left node will be 1 less than its parent, and horizontal level of the right node will be 1 more than its parent. We create a minLevel variable to store the minimum horizontal level or the leftmost node in the tree, and a maxLevel variable to store the maximum horizontal level or the rightmost node in the tree. Traverse the tree in level order and store the minLevel and maxLevel. In the end, print sum of maxLevel and absolute value of minLevel which is the vertical width of the tree.
Follow the below steps to Implement the idea:
- Initialize two variables maxLevel = 0 and minLevel = 0 and queue Q of pair of the type (Node*, Integer).
- Push (root, 0) in Q.
- Run while loop till Q is not empty
- Store the front node of the Queue in cur and the current horizontal level in count.
- If curr->left is not null then push (root->left, count-1) and update minLevel with min(minLevel, count-1).
- If curr->right is not null then push (root->right, count+1) and update maxLevel with max(maxLevel, count+1).
- Print maxLevel + abs(minLevel) + 1.
Below is the Implementation of the above approach:
C++
// C++ program to find Vertical Height of a tree
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
struct Node {
int data;
struct Node* left;
struct Node* right;
Node(int x)
{
data = x;
left = right = NULL;
}
};
int verticalWidth(Node* root)
{
// Code here
if (root == NULL)
return 0;
int maxLevel = 0, minLevel = 0;
queue<pair<Node*, int> > q;
q.push({ root, 0 }); // we take root as 0
// Level order traversal code
while (q.empty() != true) {
Node* cur = q.front().first;
int count = q.front().second;
q.pop();
if (cur->left) {
minLevel = min(minLevel,
count - 1); // as we go left,
// level is decreased
q.push({ cur->left, count - 1 });
}
if (cur->right) {
maxLevel = max(maxLevel,
count + 1); // as we go right,
// level is increased
q.push({ cur->right, count + 1 });
}
}
return maxLevel + abs(minLevel)
+ 1; // +1 for the root node, we gave it a value
// of zero
}
int main()
{
// making the tree
Node* root = new Node(7);
root->left = new Node(6);
root->right = new Node(5);
root->left->left = new Node(4);
root->left->right = new Node(3);
root->right->left = new Node(2);
root->right->right = new Node(1);
cout << "Vertical width is : " << verticalWidth(root)
<< endl;
return 0;
}
// code contributed by Anshit Bansal
// Java program to print vertical width
// of a tree
import java.util.*;
class GFG {
// A Binary Tree Node
static class Node {
int data;
Node left, right;
};
static class Pair {
Node node;
int level; // Horizontal Level
Pair(Node node, int level)
{
this.node = node;
this.level = level;
}
}
static int maxLevel = 0, minLevel = 0;
// get vertical width
static void lengthUtil(Node root)
{
Queue<Pair> q = new ArrayDeque<>();
// Adding the root node initially with level as 0;
q.add(new Pair(root, 0));
while (q.size() > 0) {
// removing the node at the peek of queue;
Pair rem = q.remove();
// If the left child of removed node exists
if (rem.node.left != null) {
// level of the left child will be 1 less
// than its parent
q.add(
new Pair(rem.node.left, rem.level - 1));
// updating the minLevel
minLevel
= Math.min(minLevel, rem.level - 1);
}
// If the right child of removed node exists
if (rem.node.right != null) {
// level of the right child will be 1 more
// than its parent
q.add(new Pair(rem.node.right,
rem.level + 1));
// updating the minLevel
maxLevel
= Math.max(minLevel, rem.level + 1);
}
}
}
static int getLength(Node root)
{
maxLevel = 0;
minLevel = 0;
lengthUtil(root);
// 1 is added to include root in the width
return (Math.abs(minLevel) + maxLevel) + 1;
}
// Utility function to create a new tree node
static Node newNode(int data)
{
Node curr = new Node();
curr.data = data;
curr.left = curr.right = null;
return curr;
}
// Driver Code
public static void main(String[] args)
{
Node root = newNode(7);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(4);
root.left.right = newNode(3);
root.right.left = newNode(2);
root.right.right = newNode(1);
System.out.println(getLength(root));
}
}
// This code is contributed by Archit Sharma
# Python code for the above approach
class Node:
# A Binary Tree Node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# get vertical width
def verticalWidth(root):
if root == None:
return 0
maxLevel = 0
minLevel = 0
q = []
# Adding the root node initially with level as 0;
q.append([root, 0]) # we take root as 0
# Level order traversal code
while len(q) != 0:
# removing the node at the peek of queue;
cur, count = q.pop(0)
# If the left child of removed node exists
if cur.left:
# updating the minLevel
minLevel = min(minLevel, count - 1)
# level of the left child will be 1 less than its parent
q.append([cur.left, count - 1])
# If the right child of removed node exists
if cur.right:
# updating the maxLevel
maxLevel = max(maxLevel, count + 1)
# level of the right child will be 1 more
# than its parent
q.append([cur.right, count + 1])
return maxLevel + abs(minLevel) + 1
# making the tree
root = Node(7)
root.left = Node(6)
root.right = Node(5)
root.left.left = Node(4)
root.left.right = Node(3)
root.right.left = Node(2)
root.right.right = Node(1)
print("Vertical width is : ", verticalWidth(root))
# This code is contributed by Potta Lokesh
// C# program to print vertical width of a tree
using System;
using System.Collections;
public class GFG {
// A Binary Tree Node
class Node {
public int data;
public Node left, right;
};
class Pair {
public Node node;
public int level; // Horizontal Level
public Pair(Node node, int level)
{
this.node = node;
this.level = level;
}
}
static int maxLevel = 0, minLevel = 0;
// get vertical width
static void lengthUtil(Node root)
{
Queue q = new Queue();
// Adding the root node initially with level as 0;
q.Enqueue(new Pair(root, 0));
while (q.Count > 0) {
// removing the node at the peek of queue;
Pair rem = (Pair)q.Dequeue();
// If the left child of removed node exists
if (rem.node.left != null) {
// level of the left child will be 1 less
// than its parent
q.Enqueue(
new Pair(rem.node.left, rem.level - 1));
// updating the minLevel
minLevel
= Math.Min(minLevel, rem.level - 1);
}
// If the right child of removed node exists
if (rem.node.right != null) {
// level of the right child will be 1 more
// than its parent
q.Enqueue(new Pair(rem.node.right,
rem.level + 1));
// updating the minLevel
maxLevel
= Math.Max(minLevel, rem.level + 1);
}
}
}
static int getLength(Node root)
{
maxLevel = 0;
minLevel = 0;
lengthUtil(root);
// 1 is added to include root in the width
return (Math.Abs(minLevel) + maxLevel) + 1;
}
// Utility function to create a new tree node
static Node newNode(int data)
{
Node curr = new Node();
curr.data = data;
curr.left = curr.right = null;
return curr;
}
static public void Main()
{
// Code
Node root = newNode(7);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(4);
root.left.right = newNode(3);
root.right.left = newNode(2);
root.right.right = newNode(1);
Console.WriteLine(getLength(root));
}
}
// This code is contributed by lokeshmvs21.
class Node {
constructor(x) {
this.data = x;
this.left = null;
this.right = null;
}
}
function verticalWidth(root) {
if (!root) {
return 0;
}
let maxLevel = 0;
let minLevel = 0;
const q = [];
q.push([root, 0]); // we take root as 0
// Level order traversal code
while (q.length !== 0) {
const [cur, count] = q.shift();
if (cur.left) {
minLevel = Math.min(minLevel, count - 1); // as we go left, level is decreased
q.push([cur.left, count - 1]);
}
if (cur.right) {
maxLevel = Math.max(maxLevel, count + 1); // as we go right, level is increased
q.push([cur.right, count + 1]);
}
}
return maxLevel + Math.abs(minLevel) + 1; // +1 for the root node, we gave it a value of zero
}
// making the tree
const root = new Node(7);
root.left = new Node(6);
root.right = new Node(5);
root.left.left = new Node(4);
root.left.right = new Node(3);
root.right.left = new Node(2);
root.right.right = new Node(1);
console.log("Vertical width is: ", verticalWidth(root));
Time Complexity: O(n). This much time is needed to traverse the tree.
Auxiliary Space: O(n). This much space is needed to maintain the queue.
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