Given a binary tree, the task is to find the bottom view of a binary tree using recursion.
Examples:
Input:
1
\
2
\
4
/ \
3 5
Output: 1 3 4 5
Input:
20
/ \
8 22
/ \ / \
5 10 21 25
/ \
9 14
Output: 5 9 21 14 25
Approach:
We can do so by using recursion and 2 arrays each with size 2n+1(for worst case), where n = number of elements in the given tree. Here, we take a Variable x which determines its Horizontal Distance. Let x is the horizontal distance of a Node. Now, the left child will have a horizontal distance of x-1(x minus 1)and the right child will have horizontal distance x+1(x plus 1). Take another Variable ‘p’ as a priority which will decide which level this element belongs to.
1 (x=0, p=0)
\
2 (x=1, p=1)
\
4 (x=2, p=2)
/ \
(x=1, p=3) 3 5 (x=3, p=3)
While Traversing the Tree In fashion Right-> Node-> Left, assign x and p to each Node and simultaneously insert the data of node in the first array if the array is empty at position (mid+x). If the array is not empty and a Node with higher Priority( p ) comes to update the array with the data of this Node as position(mid+x). The second array will be maintaining the priority( p ) of each inserted node in the first array check code for better understanding.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>using namespace std;struct Node { int data; // left and right references Node *left, *right; // Constructor of tree Node Node(int key) { data = key; left = right = NULL; }};int l = 0, r = 0;int N;// Function to generate bottom view of binary treevoid Bottom(Node* root, int arr[], int arr2[], int x, int p, int mid){ // Base case if (root == NULL) return; // To store leftmost value of x in l if (x < l) l = x; // To store rightmost value of x in r if (x > r) r = x; // To check if arr is empty at mid+x if (arr[mid + x] == INT_MIN) { // Insert data of Node at arr[mid+x] arr[mid + x] = root->data; // Insert priority of that Node at arr2[mid+x] arr2[mid + x] = p; } // If not empty and priotiy of previously inserted Node // is less than current*/ else if (arr2[mid + x] < p) { // Insert current Node data at arr[mid+x] arr[mid + x] = root->data; // Insert priotiy of that Node at arr2[mid +x] arr2[mid + x] = p; } // Go right first then left Bottom(root->right, arr, arr2, x + 1, p + 1, mid); Bottom(root->left, arr, arr2, x - 1, p + 1, mid);}// Utility function to generate bottom view of a biany treevoid bottomView(struct Node* root){ int arr[2 * N + 1]; int arr2[2 * N + 1]; for (int i = 0; i < 2 * N + 1; i++) { arr[i] = INT_MIN; arr2[i] = INT_MIN; } int mid = N, x = 0, p = 0; Bottom(root, arr, arr2, x, p, mid); for (int i = mid + l; i <= mid + r; i++) cout << arr[i] << " ";}// Driver codeint main(){ N = 5; Node* root = new Node(1); root->right = new Node(2); root->right->right = new Node(4); root->right->right->left = new Node(3); root->right->right->right = new Node(5); bottomView(root); return 0;}// This code is contributed by Sania Kumari Gupta// (kriSania804) |
C
#include <limits.h>#include <stdio.h>#include <stdlib.h>typedef struct Node { int data; struct Node *left, *right;} Node;/* Helper function that allocates a new node with the given data and NULL left and right pointers. */Node* newNode(int data){ Node* new_node = (Node*)malloc(sizeof(Node)); new_node->data = data; new_node->left = new_node->right = NULL; return new_node;}int l = 0, r = 0;int N;// Function to generate bottom view of binary treevoid Bottom(Node* root, int arr[], int arr2[], int x, int p, int mid){ // Base case if (root == NULL) return; // To store leftmost value of x in l if (x < l) l = x; // To store rightmost value of x in r if (x > r) r = x; // To check if arr is empty at mid+x if (arr[mid + x] == INT_MIN) { // Insert data of Node at arr[mid+x] arr[mid + x] = root->data; // Insert priority of that Node at arr2[mid+x] arr2[mid + x] = p; } // If not empty and priotiy of previously inserted Node // is less than current*/ else if (arr2[mid + x] < p) { // Insert current Node data at arr[mid+x] arr[mid + x] = root->data; // Insert priotiy of that Node at arr2[mid +x] arr2[mid + x] = p; } // Go right first then left Bottom(root->right, arr, arr2, x + 1, p + 1, mid); Bottom(root->left, arr, arr2, x - 1, p + 1, mid);}// Utility function to generate bottom view of a biany treevoid bottomView(struct Node* root){ int arr[2 * N + 1]; int arr2[2 * N + 1]; for (int i = 0; i < 2 * N + 1; i++) { arr[i] = INT_MIN; arr2[i] = INT_MIN; } int mid = N, x = 0, p = 0; Bottom(root, arr, arr2, x, p, mid); for (int i = mid + l; i <= mid + r; i++) printf("%d ", arr[i]);}// Driver codeint main(){ N = 5; Node* root = newNode(1); root->right = newNode(2); root->right->right = newNode(4); root->right->right->left = newNode(3); root->right->right->right = newNode(5); bottomView(root); return 0;}// This code is contributed by Sania Kumari Gupta// (kriSania804) |
Java
class GFG { static class Node { int data; // left and right references Node left, right; // Constructor of tree Node public Node(int key) { data = key; left = right = null; } }; static int l = 0, r = 0, N; // Function to generate bottom view of binary tree static void Bottom(Node root, int arr[], int arr2[], int x, int p, int mid) { // Base case if (root == null) return; // To store leftmost value of x in l if (x < l) l = x; // To store rightmost value of x in r if (x > r) r = x; // To check if arr is empty at mid+x if (arr[mid + x] == Integer.MIN_VALUE) { // Insert data of Node at arr[mid+x] arr[mid + x] = root.data; // Insert priority of that Node at arr2[mid+x] arr2[mid + x] = p; } // If not empty and priotiy of previously inserted // Node is less than current else if (arr2[mid + x] < p) { // Insert current Node data at arr[mid+x] arr[mid + x] = root.data; // Insert priotiy of that Node at arr2[mid +x] arr2[mid + x] = p; } // Go right first then left Bottom(root.right, arr, arr2, x + 1, p + 1, mid); Bottom(root.left, arr, arr2, x - 1, p + 1, mid); } // Utility function to generate bottom view of a biany // tree static void bottomView(Node root) { int[] arr = new int[2 * N + 1]; int[] arr2 = new int[2 * N + 1]; for (int i = 0; i < 2 * N + 1; i++) { arr[i] = Integer.MIN_VALUE; arr2[i] = Integer.MIN_VALUE; } int mid = N, x = 0, p = 0; Bottom(root, arr, arr2, x, p, mid); for (int i = mid + l; i <= mid + r; i++) System.out.print(arr[i] + " "); } // Driver code public static void main(String[] args) { N = 5; Node root = new Node(1); root.right = new Node(2); root.right.right = new Node(4); root.right.right.left = new Node(3); root.right.right.right = new Node(5); bottomView(root); }}// This code is contributed by Sania Kumari Gupta// (kriSania804) |
Python3
class Node: def __init__(self, data): self.data = data self.left = None self.right = Nonel = 0r = 0INT_MIN = -(2**32)# Function to generate # bottom view of # binary tree def Bottom(root, arr, arr2, x, p, mid): global INT_MIN, l, r # Base case if (root == None): return if (x < l): # To store leftmost # value of x in l l = x # To store rightmost # value of x in r if (x > r): r = x # To check if arr # is empty at mid+x if (arr[mid + x] == INT_MIN): # Insert data of Node # at arr[mid+x] arr[mid + x] = root.data # Insert priority of # that Node at arr2[mid+x] arr2[mid + x] = p # If not empty and priotiy # of previously inserted # Node is less than current*/ elif (arr2[mid + x] < p): # Insert current # Node data at arr[mid+x] arr[mid + x] = root.data # Insert priotiy of # that Node at arr2[mid +x] arr2[mid + x] = p # Go right first # then left Bottom(root.right, arr, arr2, x + 1, p + 1, mid) Bottom(root.left, arr, arr2, x - 1, p + 1, mid) # Utility function # to generate bottom # view of a biany tree def bottomView(root): global INT_MIN arr = [0]*(2 * N + 1) arr2 = [0]*(2 * N + 1) for i in range(2 * N + 1): arr[i] = INT_MIN arr2[i] = INT_MIN mid = N x = 0 p = 0 Bottom(root, arr, arr2, x, p, mid) for i in range(mid + l,mid + r + 1): print(arr[i], end = " ") # Driver code N = 5root = Node(1) root.right = Node(2) root.right.right = Node(4) root.right.right.left = Node(3) root.right.right.right = Node(5) bottomView(root) # This code is contributed by SHUBHAMSINGH10 |
C#
using System;class GFG{ class Node{ public int data;// left and right referencespublic Node left, right;// Constructor of tree Nodepublic Node(int key) { data = key; left = right = null;}};static int l = 0, r = 0, N;// Function to generate// bottom view of binary treestatic void Bottom(Node root, int []arr, int []arr2, int x, int p, int mid){ // Base case if (root == null) { return; } if (x < l) { // To store leftmost // value of x in l l = x; } // To store rightmost // value of x in r if (x > r) { r = x; } // To check if arr // is empty at mid+x if (arr[mid + x] == Int32.MinValue) { // Insert data of Node // at arr[mid+x] arr[mid + x] = root.data; // Insert priority of // that Node at arr2[mid+x] arr2[mid + x] = p; } // If not empty and priotiy // of previously inserted // Node is less than current*/ else if (arr2[mid + x] < p) { // Insert current // Node data at arr[mid+x] arr[mid + x] = root.data; // Insert priotiy of // that Node at arr2[mid +x] arr2[mid + x] = p; } // Go right first // then left Bottom(root.right, arr, arr2, x + 1, p + 1, mid); Bottom(root.left, arr, arr2, x - 1, p + 1, mid);}// Utility function to generate // bottom view of a biany treestatic void bottomView(Node root) { int[] arr = new int[2 * N + 1]; int[] arr2 = new int[2 * N + 1]; for(int i = 0; i < 2 * N + 1; i++) { arr[i] = Int32.MinValue; arr2[i] = Int32.MinValue; } int mid = N, x = 0, p = 0; Bottom(root, arr, arr2, x, p, mid); for(int i = mid + l; i <= mid + r; i++) { Console.Write(arr[i] + " "); }}// Driver codepublic static void Main(string[] args){ N = 5; Node root = new Node(1); root.right = new Node(2); root.right.right = new Node(4); root.right.right.left = new Node(3); root.right.right.right = new Node(5); bottomView(root);}}// This code is contributed by rutvik_56 |
Javascript
<script>class Node{ // Constructor of tree Node constructor(key) { this.data = key; this.left = null; this.right = null; }};var l = 0, r = 0, N;// Function to generate// bottom view of binary treefunction Bottom(root, arr, arr2, x, p, mid){ // Base case if (root == null) { return; } if (x < l) { // To store leftmost // value of x in l l = x; } // To store rightmost // value of x in r if (x > r) { r = x; } // To check if arr // is empty at mid+x if (arr[mid + x] == -1000000000) { // Insert data of Node // at arr[mid+x] arr[mid + x] = root.data; // Insert priority of // that Node at arr2[mid+x] arr2[mid + x] = p; } // If not empty and priotiy // of previously inserted // Node is less than current*/ else if (arr2[mid + x] < p) { // Insert current // Node data at arr[mid+x] arr[mid + x] = root.data; // Insert priotiy of // that Node at arr2[mid +x] arr2[mid + x] = p; } // Go right first // then left Bottom(root.right, arr, arr2, x + 1, p + 1, mid); Bottom(root.left, arr, arr2, x - 1, p + 1, mid);}// Utility function to generate // bottom view of a biany treefunction bottomView(root) { var arr = Array(2*N +1).fill(-1000000000); var arr2 = Array(2*N +1).fill(-1000000000); var mid = N, x = 0, p = 0; Bottom(root, arr, arr2, x, p, mid); for(var i = mid + l; i <= mid + r; i++) { document.write(arr[i] + " "); }}// Driver codeN = 5;var root = new Node(1);root.right = new Node(2);root.right.right = new Node(4);root.right.right.left = new Node(3);root.right.right.right = new Node(5);bottomView(root);</script> |
Output:
1 3 4 5
Time Complexity: O(N) where N is number of nodes in a given binary tree
Auxiliary space: O(n) for implicit call stack as using recursion
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