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  • Tech Mahindra aptitute test questions (17)
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Tech Mahindra Test Cases Interview Questions
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write a test case on web login and bank application and client server

60 117002

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Un-Answered Questions

what is a Depository Participant?

1826


hi my name is siraj i want to know cognos queriy language

1101


What is Material status and which view of the material master it is available?

374


At present what is the rate of excise duty on sale of Weigh Bridge Machinery to out of state from Karnataka.

1294


what is difference between marketing and kpo

1110






condition Autotrip facia come on control panel of electrical transmission line?

345


i want to run 11kv star connected motor in 6.6kv with delta connected throught inverter, is it possible?

895


Its possible to use 3core 2R of 240 sq.mm cable for 3phases & separely 50sq.mm 3core cable fully for Neutral for connection of PCC to MCC 3ph..4wire system?

926


What will happen if the voltage in my house drops from 220V to any lower voltage?

909


Difference between low impedance and high impedance Restricted Earth Fault

1775


emplementation of accounting standerds in Bangladesh, developing or copieng?

844


how you will identify the mapping level bottleneck ? if there is a mapping level bottleneck then how will you try to improve the mapping

759


How to List the Available Tables?

649


what is mean by traditional accounting ?

1189


#include #include #include #include void insert(struct btreenode **, int); void inorder(struct btreenode *); struct btreenode { struct btreenode *leftchild; struct btreenode *rightchild; int data; }; main() { struct btreenode *bt; bt=(struct btreenode *)NULL; int req,i=1,num; clrscr(); printf("Enter number of nodes"); scanf("%d",&req); while(i<=req) { printf("Enter element"); scanf("%d",&num); insert(&bt,num); i++; } inorder(bt); } void insert(struct btreenode **sr, int num) { if(*sr==NULL) { *sr=(struct btreenode *)malloc (sizeof(struct btreenode)); (*sr)->leftchild=(struct btreenode *)NULL; (*sr)->rightchild=(struct btreenode *)NULL; (*sr)->data=num; return; } else { if(num < (*sr)->data) insert(&(*sr)->leftchild,num); else insert(&(*sr)->rightchild,num); } return; } void inorder(struct btreenode *sr) { if(sr!=(struct btreenode *)NULL) { inorder(sr->leftchild); printf("\n %d",sr->data); inorder(sr->rightchild); } else return; } please Modify the given program and add two methods for post order and pre order traversals.

1418






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