Discussion on Binary Search Tree : Lowest Common Ancestor Challenge

6 years ago+ 1 comment

Easiest solution https://www.youtube.com/watch?v=N9yepbXdOkI

6 years ago+ 1 comment

Node *add1 [25]; Node *add2[25]; int i=0,j=0; Node *lca(Node *root, int v1,int v2) { // Write your code here.

    if(v1>root->data) {
        root=lca(root->right,v1,v2);
        add1[i++]=root;

    }
     else if(v2>root->data) {
        root=lca(root->right,v1,v2);
        add2[j++]=root;

    }   
    else if(v1<root->data) {
        root=lca(root->left,v1,v2);
        add1[i++]=root;

    }
     else if(v2<root->data) {
        root=lca(root->left,v1,v2);
        add2[j++]=root;

    }   
    Node *a;
    for(int i=0;i<25;i++){
        for(int j=0;j<25;j++){
            if(add1[i]==add2[2]) {a= add1[i]; return a;}
        }
    }


}

6 years ago+ 0 comments

i was basically storing the address of the of each node through which we are passing for v1 and v2 in add1 and add2 respectively
i want to ask few questions:-
solution is a structure (in this question) , when we are calling Lca function (MYTREE.LCA(ROOT,V1,V2) , does the "instance of struct solution " will have --> add1[25] , add2[25], int i,j ; when function is called
on each function call (root=lca(root->left,v1,v2) ) in each "IF" statement does the add1[25] , add2[25], int i,j , will again take memory OR one single copy of add1[25] , add2[25], int i,j ; is only there, as only one insance is made;

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