Deepak manohar's solution is an elegant one. This is very clumsy and you have a bug too.
On 10/27/06, sivaramakrishna kantharao [EMAIL PROTECTED] wrote:
hi
how abt the following one
int IsIsomorphic(tree *temp1,tree *temp2){if(temp1==NULL temp2==NULL)return 1;else if(temp1!=NULL temp2!=NULL){
great, thanks a lot this helps, i had something similar to Deepak
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would this work:
if (node1.value != node2.value)
return false;
if(node1 != null node2 != null)
{
if(node1.leftChild.value == node2.leftChild.value)
isIsomorphic(node1.leftChild, node2.leftChild)
else if(node1.leftChild.value == node2.rightChild.value)
hrmm.. so would Deepak's method work? it seems to be ok... i just dont
understand the final line return node1 == null node2 == null; why
is this needed?
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Let us say we call the following with the roots of both the treesalgo: checkIsomorphism (node1, node2){ if(node1==NULL node2==NULL) return 1; if(node1==NULL) return 0; if(node2==NULL) return 0;
child11=node1-left; child12=node1-right; child21=node2-left; child22=node2-right;
I see potential nullpointers in the code.(marked in bold)Another recursive solution:isIsomorphic(node1, node2){ if (node1 != null node2 != null) { if (node1-value != node2-value)
return 0; if (isIsomorphic(node1-left, node2-left) isIsomorphic(node1-right, node2-right)) return 1; if
