1 vote

Consider the following function height, to which pointer to the root node of a binary tree shown below is passed

Note that max(a,b) defined by #define max(a,b) (a>b)?a:b.

int height(Node *root)

The output of the above code will be _________________

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This code calculates the height of a tree where height starts from $0$

In the daigram I have written address where the respective node are stored.

$root$ stores the address $100$ i.e. points to the root node.

**For height(100)**

int height Node(*root) //root=100 ,root->left = 200 and root->right = 300 { if(!100) return -1; // if (0) so condition fails. if ( 200 && !(300)) return 1+ height(200); //if (200 && 0) = if (0) so condition fails. if ( !(200) && (300)) return 1+ height(300); //if (0 && 300) = if (0) so condition fails. return(1+max(height(200),height(300)); // this will execute }

**For height(200)**

int height Node(*root) //root=200 ,root->left = NULL and root->right = NULL { if(!200) return -1; // if (0) so condition fails. if ( NULL && !(NULL)) return 1+ height(NULL); //if (0 && 1) = if (0) so condition fails. if ( !(NULL) && NULL) return 1+ height(NULL); //if (1 && 0) = if (0) so condition fails. return(1+max(height(NULL),height(NULL)); // this will execute }

**for height (NULL)**

int height Node(*root) //root=NULL { if(!NULL) return -1; // if (!0)= if(1) so condition satisfied and it returns -1. .... .... .... }

so $height (NULL)$ returns $-1$ to $height(200)$

$height(200)$ returns returns $1+ max(-1,-1)=1-1=0$ to $height(100)$

So in $height(100)$ we get

return(1+max(0,height(300));

**For height(300)**

int height Node(*root) //root=300 ,root->left = 400 and root->right = NULL { if(!300) return -1; // if (0) so condition fails. if ( 400 && !(NULL)) return 1+ height(400); //if (1 && 1) = if (1) so condition true calls height(400) .... .... }

**For height(400)**

int height Node(*root) //root=400 ,root->left = 500 and root->right = NULL { if(!400) return -1; // if (0) so condition fails. if ( 500 && !(NULL)) return 1+ height(500); //if (1 && 1) = if (1) so condition true calls height(500) .... .... }

**For height(500)**

int height Node(*root) //root=500 ,root->left = NULL and root->right = NULL { if(!500) return -1; // if (0) so condition fails. if ( NULL && !(NULL)) return 1+ height(NULL); //if (0 && 1) = if (0) so condition fails. here 0 means false if ( !(NULL) && NULL) return 1+ height(NULL); //if (1 && 0) = if (0) so condition fails. return(1+max(height(NULL),height(NULL)); // this will execute and will call height(NULL) }

**for height (NULL)**

int height Node(*root) //root=NULL { if(!NULL) return -1; // if (!0)= if(1) so condition satisfied and it returns -1. .... .... .... }

So $height(NULL)$ returns $-1$ to $height(500)$

$height(500)$ returns $1+ max(-1,-1)=1-1=0$ to $height(400)$

$height(400)$ returns $1+0=1$ to $height(300)$

$height(300)$ returns $1+1=2$ to $height(100)$

So in $height(100)$ we get

return(1+max(0,2)); // return(1+2)= return(3)

So $3$ comes as output.

0

height(500) returns 1+max(−1,−1)=1−1=0 to height(400)

due to #define of max it will be expanded inline of code and since priority of arithmetic (+) is greater than ternary (? :)

1+max(-1,-1) ===> 1 + (-1>-1) ? -1 : -1 ======> 1+(0) ? -1 : -1 ===> 1 ? -1 : -1 ===> -1

so -1 is returned to* height(400) *

1 + (-1) = 0 is returned to *height(300)*

0 + 1 = 1 is returned to *height(100)*

also *height(200)* returns -1 to *height(100)* due to same reason

at *height(100)* :

1+max(-1,1) ===> 1 + (-1>1) ? -1 : 1 ======> 1+(0) ? -1 : 1 ===> 1 ? -1 : 1 ===> -1

following this final answer will be -1 only

@srestha @Satbir please correct if I'm wrong

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