Bottom View of a Binary Tree using Recursion

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++

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#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 tree
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] == 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 tree
void 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 code
int 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;
}

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Python3

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class Node: 
    def __init__(self, data): 
        self.data = data 
        self.left = None
        self.right = None
  
l = 0
r = 0
INT_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 =
      
    # To store rightmost 
    # value of x in r 
    if (x > r):
        r =
          
    # 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] =
          
    # 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] =
      
    # 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 = 5
root = 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

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Output:

1 3 4 5


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