Write an efficient function that takes a binary tree as input and displays the elements of tree level by level, but from last level to first. What are the time and space complexities of your solution? 

Input: 

       3

    /    \

  4      7 

 /  \     /  \

5   1  6   8 

Output: 5 1 6 8 4 7 3


code:


#include<stdio.h>
#include<conio.h>
#include<malloc.h>
#define QUEUESIZE 20

struct treenode
{
    int data;
    struct treenode *left;
    struct treenode *right;
} *root=NULL;


//queue functions
int *queue[QUEUESIZE];
int rear=-1,front=-1;

void enque(struct treenode *root)
{
    if(rear<QUEUESIZE)
    {
        queue[++rear]=root;
       
        }
        else printf("queue overflow");
    }
   
int deque()
{
    int i;
    if(front!=rear)
    i=queue[++front];
    else printf("underfolw");
    return i;
}

int isqempty()
{
    if(front==rear)
    return 1;
   
    return 0;
}


struct treenode * buildtree()
{
    struct treenode *temp1,*temp2,*temp3,*temp4,*temp5,*temp6;
     root=(struct treenode *)malloc(sizeof(struct treenode));
      temp1=(struct treenode *)malloc(sizeof(struct treenode));
       temp2=(struct treenode *)malloc(sizeof(struct treenode));
        temp3=(struct treenode *)malloc(sizeof(struct treenode));
         temp4=(struct treenode *)malloc(sizeof(struct treenode));
          temp5=(struct treenode *)malloc(sizeof(struct treenode));
           temp6=(struct treenode *)malloc(sizeof(struct treenode));
    
     root->data=4;
     root->left=temp1;
     temp1->data=5;
     root->right=temp2;
     temp2->data=6;
     temp1->left=temp3;
     temp3->data=7;
     temp3->left=temp3->right=NULL;
     temp1->right=temp4;
     temp4->data=8;
     temp4->left=NULL;
     temp4->right=NULL;
      temp2->left=temp5;
      temp5->data=6;
      temp2->right=temp6;
      temp6->data=10;
      temp5->left=temp5->right=temp6->left=temp6->right=NULL;
  printf("\n tree buildedsuccessfull"); 
       return root;
}
//no nodes
int noofnodes(struct treenode *root)
{
    if(root==NULL) return 0;
   
    return noofnodes(root->left)+noofnodes(root->right)+1;
}


//
void printfrombottam(struct treenode *root)
{
int *array[QUEUESIZE],i=0,j=0;
struct treenode *k;
array[j++]=root;
//k=noofnodes(root);
while(i<j)
{
   
    if(root->right!=NULL)
    array[j++]=root->right;
   
    if(root->left!=NULL)
    array[j++]=root->left;
   
    root=array[++i];
   
}   
for(i=j-1;i>=0;i--)
{
    k=array[i];
    printf("%d   ",k->data);
}   
}//end


//main

void main()
{
    struct treenode *root;
    int i=1;
    while(i)
    {
        printf("\n enter 1 to build elements");
        printf("\n enter 2 print elements in level wise");
        scanf("%d",&i);
   
    switch(i)
    {
    case 1:
         root=buildtree();
         break;
       
    case 2:
        printfrombottam(root);
        break;   
    }
     }
}