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Print middle level of perfect binary tree without finding height in C language
The program should print the middle level of a binary tree e.g. if there are 4 levels in a binary tree than the program must print the level 2 nodes but the demand here is to calculate the level without finding the height.
Perfect binary tree is a tree in which interior nodes must have two children and all leave nodes should be at the same level or depth.
Here,
Interior nodes 21 and 32 both are having children
Leaf nodes are 41, 59, 33 and 70 all lies at the same level.
Since it is satisfying both the properties it’s a perfect binary tree.
Example
Input : 12 21 32 41 59 33 70 Output : 21 32
The approach used here is just like finding middle elements of a linked list by checking the left and right pointer of a node i.e. whether NULL or NOT NULL by making a recursive call to a function.
The below code shows the c implementation of the algorithm given.
Algorithm
START Step 1 -> create node variable of type structure Declare int key Declare pointer of type node using *left, *right Step 2 -> create function for inserting node with parameter as value Declare temp variable of node using malloc Set temp->data = value Set temp->left = temp->right = NULL return temp step 3 - > Declare Function void middle(struct Node* a, struct Node* b) IF a = NULL||b = NULL Return IF ((b->left == NULL) && (b->right == NULL)) Print a->key Return End Call middle(a->left, b->left->left) Call middle(a->right, b->left->left) Step 4 -> Declare Function void mid_level(struct Node* node) Call middle(node, node) Step 5 -> In main() Call New passing value user want to insert as struct Node* n1 = New(13); Call mid_level(n1) STOP
Example
#include <stdio.h>
#include<stdlib.h>
struct Node {
int key;
struct Node* left, *right;
};
struct Node* New(int value) {
struct Node* temp = (struct Node*)malloc(sizeof(struct Node));
temp->key = value;
temp->left = temp->right = NULL;
return (temp);
}
void middle(struct Node* a, struct Node* b) {
if (a == NULL || b == NULL)
return;
if ((b->left == NULL) && (b->right == NULL)) {
printf("%d ",a->key);
return;
}
middle(a->left, b->left->left);
middle(a->right, b->left->left);
}
void mid_level(struct Node* node) {
middle(node, node);
}
int main() {
printf("middle level nodes are : ");
struct Node* n1 = New(13);
struct Node* n2 = New(21);
struct Node* n3 = New(44);
struct Node* n4 = New(98);
struct Node* n5 = New(57);
struct Node* n6 = New(61);
struct Node* n7 = New(70);
n2->left = n4;
n2->right = n5;
n3->left = n6;
n3->right = n7;
n1->left = n2;
n1->right = n3;
mid_level(n1);
}Output
If we run above program then it will generate following output.
middle level nodes are : 21 44
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