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Bottom View of a Binary Tree

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  • Difficulty Level : Medium
  • Last Updated : 28 Nov, 2022
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Given a Binary Tree, The task is to print the bottom view from left to right. A node x is there in output if x is the bottommost node at its horizontal distance. The horizontal distance of the left child of a node x is equal to a horizontal distance of x minus 1, and that of a right child is the horizontal distance of x plus 1. 

Examples:

Input:            20
                    /     \
                8         22
             /      \         \
          5         3        25
                   /    \      
              10       14

Output: 5, 10, 3, 14, 25.

Input:              20
                    /     \
                8         22
             /     \     /     \
          5        3  4      25
                  /    \      
              10       14
Output: 5 10 4 14 25. 
Explanation: If there are multiple bottom-most nodes for a horizontal distance from the root, then print the later one in the level traversal. 3 and 4 are both the bottom-most nodes at a horizontal distance of 0, we need to print 4. 

Recommended Practice

Bottom View of a Binary Tree Using level order traversal:

Store tree nodes in a queue for the level order traversal. Start with the horizontal distance hd as 0 of the root node, Using a Map which stores key-value pairs sorted by key and keep on adding a left child to the queue along with the horizontal distance as hd-1 and the right child as hd+1.

Every time a new horizontal distance or an existing horizontal distance is encountered put the node data for the horizontal distance as the key. For the first time it will add it to the map, next time it will replace the value. This will make sure that the bottom-most element for that horizontal distance is present on the map and if you see the tree from beneath that you will see that element. At last traverse the keys of map and print their respective values.

The following are steps to print the Bottom View of the Binary Tree. 

  1. Initialize variable hd = 0,map m with int-int key value pair and queue q to store nodes level-wise.
  2. Set root->hd = hd and push root in q
  3. Run a while loop till q is empty
    1. Store the front element in node temp and temp ->hd in variable hd and pop it then set temp->data as value for key hd in m i.e. m[hd] = temp->data. 
    2. If temp -> left is not NULL and set temp->left->hd = hd-1 as well as If temp -> right is not NULL and set temp->right->hd = hd+1 respectively. 
  4. Iterate over the keys and print the values.

Below is the implementation of the above:

C++




// C++ Program to print Bottom View of Binary Tree
#include<bits/stdc++.h>
using namespace std;
 
// Tree node class
struct Node
{
    int data; //data of the node
    int hd; //horizontal distance of the node
    Node *left, *right; //left and right references
 
    // Constructor of tree node
    Node(int key)
    {
        data = key;
        hd = INT_MAX;
        left = right = NULL;
    }
};
 
// Method that prints the bottom view.
void bottomView(Node *root)
{
    if (root == NULL)
        return;
 
    // Initialize a variable 'hd' with 0
    // for the root element.
    int hd = 0;
 
    // TreeMap which stores key value pair
    // sorted on key value
    map<int, int> m;
 
    // Queue to store tree nodes in level
    // order traversal
    queue<Node *> q;
 
    // Assign initialized horizontal distance
    // value to root node and add it to the queue.
    root->hd = hd;
    q.push(root);  // In STL, push() is used enqueue an item
 
    // Loop until the queue is empty (standard
    // level order loop)
    while (!q.empty())
    {
        Node *temp = q.front();
        q.pop();   // In STL, pop() is used dequeue an item
 
        // Extract the horizontal distance value
        // from the dequeued tree node.
        hd = temp->hd;
 
        // Put the dequeued tree node to TreeMap
        // having key as horizontal distance. Every
        // time we find a node having same horizontal
        // distance we need to replace the data in
        // the map.
        m[hd] = temp->data;
 
        // If the dequeued node has a left child, add
        // it to the queue with a horizontal distance hd-1.
        if (temp->left != NULL)
        {
            temp->left->hd = hd-1;
            q.push(temp->left);
        }
 
        // If the dequeued node has a right child, add
        // it to the queue with a horizontal distance
        // hd+1.
        if (temp->right != NULL)
        {
            temp->right->hd = hd+1;
            q.push(temp->right);
        }
    }
 
    // Traverse the map elements using the iterator.
    for (auto i = m.begin(); i != m.end(); ++i)
        cout << i->second << " ";
}
 
// Driver Code
int main()
{
    Node *root = new Node(20);
    root->left = new Node(8);
    root->right = new Node(22);
    root->left->left = new Node(5);
    root->left->right = new Node(3);
    root->right->left = new Node(4);
    root->right->right = new Node(25);
    root->left->right->left = new Node(10);
    root->left->right->right = new Node(14);
    cout << "Bottom view of the given binary tree :\n";
    bottomView(root);
    return 0;
}


Java




// Java Program to print Bottom View of Binary Tree
import java.util.*;
import java.util.Map.Entry;
 
// Tree node class
class Node
{
    int data; //data of the node
    int hd; //horizontal distance of the node
    Node left, right; //left and right references
 
    // Constructor of tree node
    public Node(int key)
    {
        data = key;
        hd = Integer.MAX_VALUE;
        left = right = null;
    }
}
 
//Tree class
class Tree
{
    Node root; //root node of tree
 
    // Default constructor
    public Tree() {}
 
    // Parameterized tree constructor
    public Tree(Node node)
    {
        root = node;
    }
 
    // Method that prints the bottom view.
    public void bottomView()
    {
        if (root == null)
            return;
 
        // Initialize a variable 'hd' with 0 for the root element.
        int hd = 0;
 
        // TreeMap which stores key value pair sorted on key value
        Map<Integer, Integer> map = new TreeMap<>();
 
         // Queue to store tree nodes in level order traversal
        Queue<Node> queue = new LinkedList<Node>();
 
        // Assign initialized horizontal distance value to root
        // node and add it to the queue.
        root.hd = hd;
        queue.add(root);
 
        // Loop until the queue is empty (standard level order loop)
        while (!queue.isEmpty())
        {
            Node temp = queue.remove();
 
            // Extract the horizontal distance value from the
            // dequeued tree node.
            hd = temp.hd;
 
            // Put the dequeued tree node to TreeMap having key
            // as horizontal distance. Every time we find a node
            // having same horizontal distance we need to replace
            // the data in the map.
            map.put(hd, temp.data);
 
            // If the dequeued node has a left child add it to the
            // queue with a horizontal distance hd-1.
            if (temp.left != null)
            {
                temp.left.hd = hd-1;
                queue.add(temp.left);
            }
            // If the dequeued node has a right child add it to the
            // queue with a horizontal distance hd+1.
            if (temp.right != null)
            {
                temp.right.hd = hd+1;
                queue.add(temp.right);
            }
        }
 
        // Extract the entries of map into a set to traverse
        // an iterator over that.
        Set<Entry<Integer, Integer>> set = map.entrySet();
 
        // Make an iterator
        Iterator<Entry<Integer, Integer>> iterator = set.iterator();
 
        // Traverse the map elements using the iterator.
        while (iterator.hasNext())
        {
            Map.Entry<Integer, Integer> me = iterator.next();
            System.out.print(me.getValue()+" ");
        }
    }
}
 
// Main driver class
public class BottomView
{
    public static void main(String[] args)
    {
        Node root = new Node(20);
        root.left = new Node(8);
        root.right = new Node(22);
        root.left.left = new Node(5);
        root.left.right = new Node(3);
        root.right.left = new Node(4);
        root.right.right = new Node(25);
        root.left.right.left = new Node(10);
        root.left.right.right = new Node(14);
        Tree tree = new Tree(root);
        System.out.println("Bottom view of the given binary tree:");
        tree.bottomView();
    }
}


Python3




# Python3 program to print Bottom
# View of Binary Tree
 
# deque supports efficient pish and pop on both ends
from collections import deque
  
# Tree node class
class Node:
     
    def __init__(self, key):
         
        self.data = key
        self.hd = float('inf')
        self.left = None
        self.right = None
  
# Method that prints the bottom view.
def bottomView(root):
 
    if (root == None):
        return
     
    # Initialize a variable 'hd' with 0
    # for the root element.
    hd = 0
     
    # Store minimum and maximum horizontal distance
    # so that we do not have to sort keys at the end
    min_hd, max_hd = 0, 0
     
    hd_dict = dict()
  
    # Queue to store tree nodes in level
    # order traversal
    q = deque()
  
    # Assign initialized horizontal distance
    # value to root node and add it to the queue.
    root.hd = hd
    q.append(root) 
  
    # Loop until the queue is empty (standard
    # level order loop)
    while q:
        curr_node = q.popleft()
          
        # Extract the horizontal distance value
        # from the dequeued tree node.
        hd = curr_node.hd
         
        # Update the minimum and maximum hd
        min_hd = min(min_hd, hd)
        max_hd = max(max_hd, hd)
  
        # Put the dequeued tree node to dictionary
        # having key as horizontal distance. Every
        # time we find a node having same horizontal
        # distance we need to update the value in
        # the map.
        hd_dict[hd] = curr_node.data
  
        # If the dequeued node has a left child, add
        # it to the queue with a horizontal distance hd-1.
        if curr_node.left:
            curr_node.left.hd = hd - 1
            q.append(curr_node.left)
  
        # If the dequeued node has a right child, add
        # it to the queue with a horizontal distance
        # hd+1.
        if curr_node.right:
            curr_node.right.hd = hd + 1
            q.append(curr_node.right)
  
    # Traverse the map from least horizontal distance to
    # most horizontal distance.
    for i in range(min_hd, max_hd+1):
        print(hd_dict[i], end = ' ')
         
# Driver Code
if __name__=='__main__':
     
    root = Node(20)
    root.left = Node(8)
    root.right = Node(22)
    root.left.left = Node(5)
    root.left.right = Node(3)
    root.right.left = Node(4)
    root.right.right = Node(25)
    root.left.right.left = Node(10)
    root.left.right.right = Node(14)
     
    print("Bottom view of the given binary tree :")
     
    bottomView(root)
     
# This code is contributed by rutvik_56


C#




// C# program to print Bottom View of Binary Tree
using System;
using System.Collections;
using System.Collections.Generic;
  
// Tree node class
class Node
{
     
    // Data of the node
    public int data;
     
    // Horizontal distance of the node
    public int hd;
     
    // left and right references
    public Node left, right;
     
    // Constructor of tree node
    public Node(int key)
    {
        data = key;
        hd = 1000000;
        left = right = null;
    }
}
 
// Tree class
class Tree
{
     
    // Root node of tree
    Node root;
     
    // Default constructor
    public Tree(){}
     
    // Parameterized tree constructor
    public Tree(Node node)
    {
        root = node;
    }
  
    // Method that prints the bottom view.
    public void bottomView()
    {
        if (root == null)
            return;
             
        // Initialize a variable 'hd' with
        // 0 for the root element.
        int hd = 0;
  
        // TreeMap which stores key value
        // pair sorted on key value
        SortedDictionary<int,
                         int> map = new SortedDictionary<int,
                                                         int>();
  
        // Queue to store tree nodes in level order
        // traversal
        Queue queue = new Queue();
         
        // Assign initialized horizontal distance
        // value to root node and add it to the queue.
        root.hd = hd;
        queue.Enqueue(root);
  
        // Loop until the queue is empty
        // (standard level order loop)
        while (queue.Count != 0)
        {
            Node temp = (Node) queue.Dequeue();
  
            // Extract the horizontal distance value
            // from the dequeued tree node.
            hd = temp.hd;
  
            // Put the dequeued tree node to TreeMap
            // having key as horizontal distance.
            // Every time we find a node having same
            // horizontal distance we need to replace
            // the data in the map.
            map[hd] = temp.data;
  
            // If the dequeued node has a left child
            // add it to the queue with a horizontal
            // distance hd-1.
            if (temp.left != null)
            {
                temp.left.hd = hd - 1;
                queue.Enqueue(temp.left);
            }
             
            // If the dequeued node has a right
            // child add it to the queue with a
            // horizontal distance hd+1.
            if (temp.right != null)
            {
                temp.right.hd = hd + 1;
                queue.Enqueue(temp.right);
            }
        }
         
        foreach(int i in map.Values)
        {
            Console.Write(i + " ");
        }
    }
}
  
public class BottomView{
   
// Driver code
public static void Main(string[] args)
{
    Node root = new Node(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.left.left = new Node(5);
    root.left.right = new Node(3);
    root.right.left = new Node(4);
    root.right.right = new Node(25);
    root.left.right.left = new Node(10);
    root.left.right.right = new Node(14);
    Tree tree = new Tree(root);
     
    Console.WriteLine("Bottom view of the " +
                      "given binary tree:");
     
    tree.bottomView();
}
}
 
// This code is contributed by pratham76


Javascript




<script>
 
      // JavaScript program to print Bottom View of Binary Tree
      // Tree node class
      class Node {
        // Constructor of tree node
        constructor(key) {
          this.data = key; // Data of the node
          this.hd = 1000000; // Horizontal distance of the node
          this.left = null; // left and right references
          this.right = null;
        }
      }
 
      // Tree class
      class Tree {
        // Parameterized tree constructor
        constructor(node) {
          // Root node of tree
          this.root = node;
        }
 
        // Method that prints the bottom view.
        bottomView() {
          if (this.root == null) return;
 
          // Initialize a variable 'hd' with
          // 0 for the root element.
          var hd = 0;
 
          // TreeMap which stores key value
          // pair sorted on key value
          var map = {};
 
          // Queue to store tree nodes in level order
          // traversal
          var queue = [];
 
          // Assign initialized horizontal distance
          // value to root node and add it to the queue.
          this.root.hd = hd;
          queue.push(this.root);
 
          // Loop until the queue is empty
          // (standard level order loop)
          while (queue.length != 0) {
            var temp = queue.shift();
 
            // Extract the horizontal distance value
            // from the dequeued tree node.
            hd = temp.hd;
 
            // Put the dequeued tree node to TreeMap
            // having key as horizontal distance.
            // Every time we find a node having same
            // horizontal distance we need to replace
            // the data in the map.
            map[hd] = temp.data;
 
            // If the dequeued node has a left child
            // add it to the queue with a horizontal
            // distance hd-1.
            if (temp.left != null) {
              temp.left.hd = hd - 1;
              queue.push(temp.left);
            }
 
            // If the dequeued node has a right
            // child add it to the queue with a
            // horizontal distance hd+1.
            if (temp.right != null) {
              temp.right.hd = hd + 1;
              queue.push(temp.right);
            }
          }
 
          for (const [key, value] of Object.entries(map).sort(
            (a, b) => a[0] - b[0]
          )) {
            document.write(value + " ");
          }
        }
      }
 
      // Driver code
      var root = new Node(20);
      root.left = new Node(8);
      root.right = new Node(22);
      root.left.left = new Node(5);
      root.left.right = new Node(3);
      root.right.left = new Node(4);
      root.right.right = new Node(25);
      root.left.right.left = new Node(10);
      root.left.right.right = new Node(14);
      var tree = new Tree(root);
 
      document.write("Bottom view of the " + "given binary tree:<br>");
 
      tree.bottomView();
       
</script>


Output

Bottom view of the given binary tree :
5 10 4 14 25 

Time Complexity: O(N * logN)
Auxiliary Space: O(N)

Bottom View of a Binary Tree Using Depth first search:

Create a map where the key is the horizontal distance and the value is a pair(a, b) where a is the value of the node and b is the height of the node. Perform a pre-order traversal of the tree. If the current node at a horizontal distance of h is the first we’ve seen, insert it into the map. Otherwise, compare the node with the existing one in map and if the height of the new node is greater, update the Map.

Follow the below steps to implement the idea:

Create a map where the key is the horizontal distance and the value is a pair(a, b) where a is the value of the node and b is the height of the node. 

  • Recursively apply Depth first search on the Tree starting from root node and keep a track of height curr and horizontal distance as hd of the current node with respect to root node as root node’s height is 0.
    • If there’s no node as value for horizontal distance hd in map then set map[hd] = make_pair(root -> data, curr).
    • Else store the pair value of mp[hd] in p. Then 
      • If p.second <= curr then set m[hd].second = curr and m[hd].first = root -> data
    • Call depth first search for left node of root node with height curr + 1 and horizontal distance as hd – 1 and right node of root node with height curr + 1 and horizontal distance as hd + 1
    • Base case would be to return when root is NULL.

Below is the implementation of the above approach.

C++




// C++ Program to print Bottom View of Binary Tree
#include <bits/stdc++.h>
#include <map>
using namespace std;
 
// Tree node class
struct Node
{
    // data of the node
    int data;
     
    // horizontal distance of the node
    int hd;
     
    //left and right references
    Node * left, * right;
     
    // Constructor of tree node
    Node(int key)
    {
        data = key;
        hd = INT_MAX;
        left = right = NULL;
    }
};
 
void printBottomViewUtil(Node * root, int curr, int hd, map <int, pair <int, int>> & m)
{
    // Base case
    if (root == NULL)
        return;
     
    // If node for a particular
    // horizontal distance is not
    // present, add to the map.
    if (m.find(hd) == m.end())
    {
        m[hd] = make_pair(root -> data, curr);
    }
    // Compare height for already
    // present node at similar horizontal
    // distance
    else
    {
        pair < int, int > p = m[hd];
        if (p.second <= curr)
        {
            m[hd].second = curr;
            m[hd].first = root -> data;
        }
    }
     
    // Recur for left subtree
    printBottomViewUtil(root -> left, curr + 1, hd - 1, m);
     
    // Recur for right subtree
    printBottomViewUtil(root -> right, curr + 1, hd + 1, m);
}
 
void printBottomView(Node * root)
{
     
    // Map to store Horizontal Distance,
    // Height and Data.
    map < int, pair < int, int > > m;
     
    printBottomViewUtil(root, 0, 0, m);
     
     // Prints the values stored by printBottomViewUtil()
    map < int, pair < int, int > > ::iterator it;
    for (it = m.begin(); it != m.end(); ++it)
    {
        pair < int, int > p = it -> second;
        cout << p.first << " ";
    }
}
 
int main()
{
    Node * root = new Node(20);
    root -> left = new Node(8);
    root -> right = new Node(22);
    root -> left -> left = new Node(5);
    root -> left -> right = new Node(3);
    root -> right -> left = new Node(4);
    root -> right -> right = new Node(25);
    root -> left -> right -> left = new Node(10);
    root -> left -> right -> right = new Node(14);
    cout << "Bottom view of the given binary tree :\n";
    printBottomView(root);
    return 0;
}


Java




// Java program to print Bottom View of Binary Tree
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Tree node class
static class Node
{
     
    // Data of the node
    int data;
 
    // Horizontal distance of the node
    int hd;
 
    // Left and right references
    Node left, right;
 
    // Constructor of tree node
    public Node(int key)
    {
        data = key;
        hd = Integer.MAX_VALUE;
        left = right = null;
    }
}
 
static void printBottomViewUtil(Node root, int curr, int hd,
                                TreeMap<Integer, int[]> m)
{
     
    // Base case
    if (root == null)
        return;
 
    // If node for a particular
    // horizontal distance is not
    // present, add to the map.
    if (!m.containsKey(hd))
    {
        m.put(hd, new int[]{ root.data, curr });
    }
     
    // Compare height for already
    // present node at similar horizontal
    // distance
    else
    {
        int[] p = m.get(hd);
        if (p[1] <= curr)
        {
            p[1] = curr;
            p[0] = root.data;
        }
        m.put(hd, p);
    }
 
    // Recur for left subtree
    printBottomViewUtil(root.left, curr + 1,
                        hd - 1, m);
 
    // Recur for right subtree
    printBottomViewUtil(root.right, curr + 1,
                        hd + 1, m);
}
 
static void printBottomView(Node root)
{
 
    // Map to store Horizontal Distance,
    // Height and Data.
    TreeMap<Integer, int[]> m = new TreeMap<>();
 
    printBottomViewUtil(root, 0, 0, m);
 
    // Prints the values stored by printBottomViewUtil()
    for(int val[] : m.values())
    {
        System.out.print(val[0] + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    Node root = new Node(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.left.left = new Node(5);
    root.left.right = new Node(3);
    root.right.left = new Node(4);
    root.right.right = new Node(25);
    root.left.right.left = new Node(10);
    root.left.right.right = new Node(14);
 
    System.out.println(
        "Bottom view of the given binary tree:");
 
    printBottomView(root);
}
}
 
// This code is contributed by Kingash


Python3




# Python3 program to print Bottom
# View of Binary Tree
class Node:
     
    def __init__(self, key = None,
                      left = None,
                     right = None):
                          
        self.data = key
        self.left = left
        self.right = right
         
def printBottomView(root):
     
      # Create a dictionary where
    # key -> relative horizontal distance
    # of the node from root node and
    # value -> pair containing node's
    # value and its level
    d = dict()
     
    printBottomViewUtil(root, d, 0, 0)
     
    # Traverse the dictionary in sorted
    # order of their keys and print
    # the bottom view
    for i in sorted(d.keys()):
        print(d[i][0], end = " ")
 
def printBottomViewUtil(root, d, hd, level):
     
      # Base case
    if root is None:
        return
     
    # If current level is more than or equal
    # to maximum level seen so far for the
    # same horizontal distance or horizontal
    # distance is seen for the first time,
    # update the dictionary
    if hd in d:
        if level >= d[hd][1]:
            d[hd] = [root.data, level]
    else:
        d[hd] = [root.data, level]
         
    # recur for left subtree by decreasing
    # horizontal distance and increasing
    # level by 1
    printBottomViewUtil(root.left, d, hd - 1,
                                   level + 1)
     
    # recur for right subtree by increasing
    # horizontal distance and increasing
    # level by 1
    printBottomViewUtil(root.right, d, hd + 1,
                                    level + 1)
 
# Driver Code   
if __name__ == '__main__':
     
    root = Node(20)
    root.left = Node(8)
    root.right = Node(22)
    root.left.left = Node(5)
    root.left.right = Node(3)
    root.right.left = Node(4)
    root.right.right = Node(25)
    root.left.right.left = Node(10)
    root.left.right.right = Node(14)
     
    print("Bottom view of the given binary tree :")
     
    printBottomView(root)
 
# This code is contributed by tusharroy


C#




// C# program to print Bottom View of Binary Tree
using System;
using System.Collections;
using System.Collections.Generic;
 
// Tree node class
public class Node {
    // data of the node
    public int data;
 
    // horizontal distance of the node
    public int hd;
 
    // left and right references
    public Node left, right;
 
    // Constructor of tree node
    public Node(int key)
    {
        data = key;
        hd = Int32.MaxValue;
        left = right = null;
    }
}
 
public class BinaryTree {
    public static void printBottomViewUtil(
        Node root, int curr, int hd,
        SortedDictionary<int, Tuple<int, int> > m)
    {
        // Base case
        if (root == null)
            return;
 
        // If node for a particular
        // horizontal distance is not
        // present, add to the map.
        if (!m.ContainsKey(hd)) {
            m[hd] = Tuple.Create(root.data, curr);
        }
 
        // Compare height for already
        // present node at similar horizontal
        // distance
        else {
            Tuple<int, int> p = m[hd];
            if (p.Item2 <= curr) {
                p = Tuple.Create(root.data, curr);
            }
            m[hd] = p;
        }
 
        // Recur for left subtree
        printBottomViewUtil(root.left, curr + 1, hd - 1, m);
 
        // Recur for right subtree
        printBottomViewUtil(root.right, curr + 1, hd + 1,
                            m);
    }
    public static void printBottomView(Node root)
    {
        // Map to store Horizontal Distance,
        // Height and Data.
        SortedDictionary<int, Tuple<int, int> > m
            = new SortedDictionary<int, Tuple<int, int> >();
        printBottomViewUtil(root, 0, 0, m);
 
        // Prints the values stored by printBottomViewUtil()
        foreach(KeyValuePair<int, Tuple<int, int> > it in m)
        {
            Tuple<int, int> p = it.Value;
            Console.Write(p.Item1 + " ");
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        Node root = new Node(20);
        root.left = new Node(8);
        root.right = new Node(22);
        root.left.left = new Node(5);
        root.left.right = new Node(3);
        root.right.left = new Node(4);
        root.right.right = new Node(25);
        root.left.right.left = new Node(10);
        root.left.right.right = new Node(14);
        Console.WriteLine(
            "Bottom view of the given binary tree :");
        printBottomView(root);
    }
}
 
// This code is contributed by Tapesh (tapeshdua420)


Output

Bottom view of the given binary tree :
5 10 4 14 25 

Time Complexity: O(N * logN) the extra logN factor is for accessing map
Auxiliary Space: O(N)

Bottom View of a Binary Tree Using unordered_map O(n):

Create a unordered_map where the key is the horizontal distance and the value is a int x, where x is the value of the node .Perform a level order traversal of the tree. For every node at a horizontal distance of h we will store its value in unordered_map<int,int>  (eg <vertical_index , root->data> ). 
suppose on a X plane.
  …  -2    -1     0     1    2 ……

        .        .     20   .      .
        .        .  /     \  .      .
        .        8         22     .
        .    /   .    \      .  \    .
         5      .    3      .   25
                 .  /    \   .   
                10       14
Now we will perform level order traversal and for every node we visit we will store its vertical index and withs value.

C++




#include <bits/stdc++.h>
using namespace std;
 
// Tree node class
struct Node
{
    // data of the node
    int data;
    // horizontal distance of the node
    int hd;
    //left and right references
    Node * left, * right;
    // Constructor of tree node
    Node(int key)
    {
        data = key;
        hd = INT_MAX;
        left = right = NULL;
    }
};
 
 
void printBottomView(Node * root)
{
    if(!root) return ;//if root is NULL
    unordered_map<int,int> hash; // <vertical_index , root->data>
    int leftmost = 0; // to store the leftmost index so that we move from left to right
    queue<pair<Node*,int>> q; // pair<Node*,vertical Index>  for level order traversal.
    q.push({root,0}); // push the root and 0 vertial index
    while(!q.empty()){
        auto top = q.front(); //  store q.front() in top variable
        q.pop();
        Node *temp = top.first; // store the Node in temp for left and right nodes
        int ind = top.second; // store the vertical index of current node
        hash[ind] = temp->data; // store the latest vertical_index(key) -> root->data(value)
        leftmost = min(ind,leftmost); // have the leftmost vertical index
        if(temp->left){ q.push({temp->left,ind-1});}// check if any node of left then go in negative direction
        if(temp->right){ q.push({temp->right,ind+1});} //check if any node of left then go in positive direction
    }
    //teverse each value in hash from leftmost to positive side till key is available
    while(hash.find(leftmost)!=hash.end()){ cout<<hash[leftmost++]<<" "; }
}
 
int main()
{
    Node * root = new Node(20);
    root -> left = new Node(8);
    root -> right = new Node(22);
    root -> left -> left = new Node(5);
    root -> left -> right = new Node(3);
    root -> right -> left = new Node(4);
    root -> right -> right = new Node(25);
    root -> left -> right -> left = new Node(10);
    root -> left -> right -> right = new Node(14);
    cout << "Bottom view of the given binary tree :\n";
    printBottomView(root);
    return 0;
}


Java




// Java implementation of the above approach
import java.util.*;
 
class gfg2
{
 
  // Tree node class
  static class Node
  {
 
    // data of the node
    int data;
 
    // horizontal distance of the node
    int hd;
 
    // left and right references
    Node left, right;
 
    // Constructor of tree node
    Node(int key)
    {
      data = key;
      hd = Integer.MAX_VALUE;
      left = right = null;
    }
  }
 
  static class pair {
    Node first;
    int second;
    pair(Node f, int s)
    {
      first = f;
      second = s;
    }
  }
 
  static void printBottomView(Node root)
  {
    if (root == null)
      return; // if root is NULL
    HashMap<Integer, Integer> hash
      = new HashMap<>(); // <vertical_index ,
    // root->data>
    int leftmost = 0; // to store the leftmost index so
    // that we move from left to right
    Queue<pair> q
      = new ArrayDeque<>(); // pair<Node*,vertical
    // Index>  for level order
    // traversal.
 
    q.add(new pair(
      root, 0)); // push the root and 0 vertial index
    while (!q.isEmpty()) {
      pair top = q.peek(); //  store q.front() in top
      //  variable
      q.remove();
      Node temp
        = top.first; // store the Node in temp for
      // left and right nodes
      int ind = top.second; // store the vertical
      // index of current node
      hash.put(ind,
               temp.data); // store the latest
      // vertical_index(key) ->
      // root->data(value)
      leftmost = Math.min(
        ind, leftmost); // have the leftmost
      // vertical index
      if (temp.left != null) {
        q.add(new pair(temp.left, ind - 1));
      } // check if any node of left then go in
      // negative direction
      if (temp.right != null) {
        q.add(new pair(temp.right, ind + 1));
      } // check if any node of left then go in
      // positive direction
    }
    // teverse each value in hash from leftmost to
    // positive side till key is available
    while (hash.containsKey(leftmost)) {
      System.out.print(
        hash.getOrDefault(leftmost++, 0) + " ");
    }
  }
 
  public static void main(String[] args)
  {
    Node root = new Node(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.left.left = new Node(5);
    root.left.right = new Node(3);
    root.right.left = new Node(4);
    root.right.right = new Node(25);
    root.left.right.left = new Node(10);
    root.left.right.right = new Node(14);
    System.out.println(
      "Bottom view of the given binary tree :");
    printBottomView(root);
  }
}
 
// This code is contributed by karandeep1234


Output

Bottom view of the given binary tree :
5 10 4 14 25 

Time Complexity: O(N )
Auxiliary Space: O(N)


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