Bottom View of a Binary Tree
Given a Binary Tree, we need 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. Horizontal distance of left child of a node x is equal to horizontal distance of x minus 1, and that of right child is horizontal distance of x plus 1.
Examples:
20
/ \
8 22
/ \ \
5 3 25
/ \
10 14For the above tree the output should be 5, 10, 3, 14, 25.
If there are multiple bottom-most nodes for a horizontal distance from root, then print the later one in level traversal. For example, in the below diagram, 3 and 4 are both the bottom-most nodes at horizontal distance 0, we need to print 4.
20
/ \
8 22
/ \ / \
5 3 4 25
/ \
10 14For the above tree the output should be 5, 10, 4, 14, 25.
Method 1 – Using Queue
The following are steps to print Bottom View of Binary Tree.
1. We put tree nodes in a queue for the level order traversal.
2. Start with the horizontal distance(hd) 0 of the root node, keep on adding left child to queue along with the horizontal distance as hd-1 and right child as hd+1.
3. Also, use a TreeMap which stores key value pair sorted on key.
4. Every time, we encounter a new horizontal distance or an existing horizontal distance put the node data for the horizontal distance as key. For the first time it will add 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 in the map and if you see the tree from beneath that you will see that element.
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 classstruct 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 Codeint 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 Treeimport java.util.*;import java.util.Map.Entry;// Tree node classclass 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 classclass 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 classpublic 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 # Tree node classclass Node: def __init__(self, key): self.data = key self.hd = 1000000 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 # TreeMap which stores key value pair # sorted on key value m = dict() # Queue to store tree nodes in level # order traversal q = [] # Assign initialized horizontal distance # value to root node and add it to the queue. root.hd = hd # In STL, append() is used enqueue an item q.append(root) # Loop until the queue is empty (standard # level order loop) while (len(q) != 0): temp = q[0] # In STL, pop() is used dequeue an item q.pop(0) # 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 != None): temp.left.hd = hd - 1 q.append(temp.left) # If the dequeued node has a right child, add # it to the queue with a horizontal distance # hd+1. if (temp.right != None): temp.right.hd = hd + 1 q.append(temp.right) # Traverse the map elements using the iterator. for i in sorted(m.keys()): print(m[i], end = ' ') # Driver Codeif __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 Treeusing System;using System.Collections;using System.Collections.Generic; // Tree node classclass 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 classclass 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 codepublic 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
Method 2- Using HashMap()
This method is contributed by Ekta Goel.
Approach:
Create a map like, map where key is the horizontal distance and 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 in the map. Otherwise, compare the node with the existing one in map and if the height of the new node is greater, update in the Map.
Below is the implementation of the above:
C++
// C++ Program to print Bottom View of Binary Tree#include <bits/stdc++.h>#include <map>using namespace std;// Tree node classstruct 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 Treeimport java.io.*;import java.lang.*;import java.util.*;class GFG{// Tree node classstatic 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 Codepublic 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 Treeclass 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 |
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